119888
A point \(P\) lies on the circle \(x^2+y^2=169\). If \(Q=\) \((5,12)\) and \(R=(-12,5)\), then the angle \(\angle Q P R\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, circle \(\mathrm{x}^2+\mathrm{y}^2=169\) Let point \(\mathrm{p}(\mathrm{x}, \mathrm{y}), \mathrm{q}=(5,12), \mathrm{R}=(-12,5)\) Now, points \(\mathrm{Q}\) and \(\mathrm{R}\) also lies on the circle Slope of line \(\mathrm{OR}=\frac{12}{5}=\mathrm{m}_1\) Slope of line \(\mathrm{OR}=\frac{-5}{12}=\mathrm{m}_2\) We know that, Angle suspended by any chord in circle to circumference is half of angle formed at centre So, \(\quad \angle \mathrm{QOR}=\frac{\pi}{2}\) So, \(\quad \angle \mathrm{QPR}=\frac{1}{2} \angle \mathrm{QOR}=\frac{\pi}{4}\)
WB JEE-2013
Conic Section
119889
If the lengths of the tangents from the point \((1\), 2) to the circles \(x^2+y^2+x+y-4=0\) and \(3 x^2+\) \(3 y^2-x-y-\lambda=0\) are in the ratio \(4: 3\), then the value of \(\lambda\) is
1 \(\frac{21}{2}\)
2 \(\frac{21}{4}\)
3 \(\frac{21}{5}\)
4 \(\frac{21}{11}\)
Explanation:
B Let \(\mathrm{S}_1\) and \(\mathrm{S}_2\) be the two circles given that. \(S_1 =x^2+y^2+x+y-4=0\) \(S_2 =3 x^2+3 y^2-x-y-\lambda=0\) \(=x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0\) Length of a tangent drawn from a point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to a circle \(x^2+y^2+2 g x+2 f y+c=0\) Length of tangent \(L=\sqrt{x_1{ }^2+y_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let \(\mathrm{L}_1\) and \(\mathrm{L}_2\) be the length of tangent drawn from \((1,2)\) \(\text { Given, } \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{4}{3}=\frac{\sqrt{1^2+2^2+1+2-4}}{\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}}\) \(\frac{16}{9}=\frac{4}{4-\frac{\lambda}{3}}\) \(16-\frac{4 \lambda}{3}=9\) \(48-4 \lambda=27\) \(4 \lambda=48-27\) \(4 \lambda=21\) \(\lambda=\frac{21}{4}\)
Jamia Millia Islamia-2010
Conic Section
119891
Suppose the tangents drawn to the circle \(x^2+y^2\) \(-6 x-4 y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P, A\) and \(B\) is
1 \((2,5)\)
2 \((-2,-5)\)
3 \((-2,5)\)
4 \((2,-5)\)
Explanation:
A Given equation of circle. \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-4 \mathrm{y}-11=0\) Equation of \(\mathrm{AB}\) of chord of contact of \(\mathrm{P}\) is using general equation of circle is \(\mathrm{x}_1 \mathrm{x}+\mathrm{y}_1 \mathrm{y}+\mathrm{g}\left(\mathrm{x}_1+\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}_1+\mathrm{y}\right)+\mathrm{c}=0\) we get, at point \((1,8)\) \(\begin{aligned} & 1 x+8 y-3(x+1)-2(y+8)-11=0 \\ & -2 x+6 y-30=0 \\ & x-3 y+15=0 \end{aligned}\) Equation of any circle through \(A B\) is \(x^2+y^2-6 x-4 y-11+\lambda(x-3 y+15)=0\) It will pass through \(P(1,8)\) if \(\begin{aligned} & 1+64-6-32-11+\lambda(1-24+15)=0 \\ & 16-8 \lambda=0 \\ & \lambda=2 \end{aligned}\) By putting the value of \(\lambda\) in equation (i) \(\begin{aligned} & x^2+y^2-6 x-4 y-11+2 x-6 y+30=0 \\ & x^2+y^2-4 x-10 y+19=0 \end{aligned}\) Thus centre of given circle is \((-g,-f)=(2,5)\)
AP EAMCET-06.07.2022
Conic Section
119892
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
1 \(\left(\frac{-3}{2}, 0\right)\)
2 \(\left(\frac{-5}{2}, 2\right)\)
3 \(\left(\frac{-3}{2}, \frac{5}{2}\right)\)
4 \((-4,0)\)
Explanation:
D Since circle touching y-axis at \((0,2)\) is \((x-0)^2+(y-2)^2+\lambda x=0\) passes through \((-1,0)\) \(\therefore 1+4-\lambda=0\) \(\lambda=5\) \(\therefore \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-4 \mathrm{y}+4=0\) Put value of option in equation of circle we get, circle passes through \((-4,0)\)
AP EAMCET-06.07.2022
Conic Section
119893
The slop of the normal to the circle \(x^2+y^2+\) \(\mathbf{2 g x}+\mathbf{2 f y}+\mathbf{c}=\mathbf{0}\) at \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is
1 \(-\left(\frac{x_1+g}{y_1+f}\right)\)
2 \(-\left(\frac{y_1+f}{x_1+g}\right)\)
3 \(\frac{x_1+g}{y_1+f}\)
4 \(\frac{y_1+f}{x_1+g}\)
Explanation:
D Given circle equation \(x^2+y^2+2 g x+2 f y+c=0 \text { at }\left(x_1, y_1\right)\) \(\text { If we know the slope of the tangent we can find the }\) \(\text { equation of tangent at }\left(x_1, y_2\right)\) Tangent and the radius \(\mathrm{CP}\) are perpendicular to each other Slope of tangent \(\times\) slope of \(\mathrm{cp}=-1\) \(\text { slope of tangent }=\frac{-1}{\text { slope of } \mathrm{CP}}\)slope of \(\mathrm{CP}=\frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}\)
119888
A point \(P\) lies on the circle \(x^2+y^2=169\). If \(Q=\) \((5,12)\) and \(R=(-12,5)\), then the angle \(\angle Q P R\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, circle \(\mathrm{x}^2+\mathrm{y}^2=169\) Let point \(\mathrm{p}(\mathrm{x}, \mathrm{y}), \mathrm{q}=(5,12), \mathrm{R}=(-12,5)\) Now, points \(\mathrm{Q}\) and \(\mathrm{R}\) also lies on the circle Slope of line \(\mathrm{OR}=\frac{12}{5}=\mathrm{m}_1\) Slope of line \(\mathrm{OR}=\frac{-5}{12}=\mathrm{m}_2\) We know that, Angle suspended by any chord in circle to circumference is half of angle formed at centre So, \(\quad \angle \mathrm{QOR}=\frac{\pi}{2}\) So, \(\quad \angle \mathrm{QPR}=\frac{1}{2} \angle \mathrm{QOR}=\frac{\pi}{4}\)
WB JEE-2013
Conic Section
119889
If the lengths of the tangents from the point \((1\), 2) to the circles \(x^2+y^2+x+y-4=0\) and \(3 x^2+\) \(3 y^2-x-y-\lambda=0\) are in the ratio \(4: 3\), then the value of \(\lambda\) is
1 \(\frac{21}{2}\)
2 \(\frac{21}{4}\)
3 \(\frac{21}{5}\)
4 \(\frac{21}{11}\)
Explanation:
B Let \(\mathrm{S}_1\) and \(\mathrm{S}_2\) be the two circles given that. \(S_1 =x^2+y^2+x+y-4=0\) \(S_2 =3 x^2+3 y^2-x-y-\lambda=0\) \(=x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0\) Length of a tangent drawn from a point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to a circle \(x^2+y^2+2 g x+2 f y+c=0\) Length of tangent \(L=\sqrt{x_1{ }^2+y_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let \(\mathrm{L}_1\) and \(\mathrm{L}_2\) be the length of tangent drawn from \((1,2)\) \(\text { Given, } \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{4}{3}=\frac{\sqrt{1^2+2^2+1+2-4}}{\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}}\) \(\frac{16}{9}=\frac{4}{4-\frac{\lambda}{3}}\) \(16-\frac{4 \lambda}{3}=9\) \(48-4 \lambda=27\) \(4 \lambda=48-27\) \(4 \lambda=21\) \(\lambda=\frac{21}{4}\)
Jamia Millia Islamia-2010
Conic Section
119891
Suppose the tangents drawn to the circle \(x^2+y^2\) \(-6 x-4 y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P, A\) and \(B\) is
1 \((2,5)\)
2 \((-2,-5)\)
3 \((-2,5)\)
4 \((2,-5)\)
Explanation:
A Given equation of circle. \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-4 \mathrm{y}-11=0\) Equation of \(\mathrm{AB}\) of chord of contact of \(\mathrm{P}\) is using general equation of circle is \(\mathrm{x}_1 \mathrm{x}+\mathrm{y}_1 \mathrm{y}+\mathrm{g}\left(\mathrm{x}_1+\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}_1+\mathrm{y}\right)+\mathrm{c}=0\) we get, at point \((1,8)\) \(\begin{aligned} & 1 x+8 y-3(x+1)-2(y+8)-11=0 \\ & -2 x+6 y-30=0 \\ & x-3 y+15=0 \end{aligned}\) Equation of any circle through \(A B\) is \(x^2+y^2-6 x-4 y-11+\lambda(x-3 y+15)=0\) It will pass through \(P(1,8)\) if \(\begin{aligned} & 1+64-6-32-11+\lambda(1-24+15)=0 \\ & 16-8 \lambda=0 \\ & \lambda=2 \end{aligned}\) By putting the value of \(\lambda\) in equation (i) \(\begin{aligned} & x^2+y^2-6 x-4 y-11+2 x-6 y+30=0 \\ & x^2+y^2-4 x-10 y+19=0 \end{aligned}\) Thus centre of given circle is \((-g,-f)=(2,5)\)
AP EAMCET-06.07.2022
Conic Section
119892
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
1 \(\left(\frac{-3}{2}, 0\right)\)
2 \(\left(\frac{-5}{2}, 2\right)\)
3 \(\left(\frac{-3}{2}, \frac{5}{2}\right)\)
4 \((-4,0)\)
Explanation:
D Since circle touching y-axis at \((0,2)\) is \((x-0)^2+(y-2)^2+\lambda x=0\) passes through \((-1,0)\) \(\therefore 1+4-\lambda=0\) \(\lambda=5\) \(\therefore \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-4 \mathrm{y}+4=0\) Put value of option in equation of circle we get, circle passes through \((-4,0)\)
AP EAMCET-06.07.2022
Conic Section
119893
The slop of the normal to the circle \(x^2+y^2+\) \(\mathbf{2 g x}+\mathbf{2 f y}+\mathbf{c}=\mathbf{0}\) at \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is
1 \(-\left(\frac{x_1+g}{y_1+f}\right)\)
2 \(-\left(\frac{y_1+f}{x_1+g}\right)\)
3 \(\frac{x_1+g}{y_1+f}\)
4 \(\frac{y_1+f}{x_1+g}\)
Explanation:
D Given circle equation \(x^2+y^2+2 g x+2 f y+c=0 \text { at }\left(x_1, y_1\right)\) \(\text { If we know the slope of the tangent we can find the }\) \(\text { equation of tangent at }\left(x_1, y_2\right)\) Tangent and the radius \(\mathrm{CP}\) are perpendicular to each other Slope of tangent \(\times\) slope of \(\mathrm{cp}=-1\) \(\text { slope of tangent }=\frac{-1}{\text { slope of } \mathrm{CP}}\)slope of \(\mathrm{CP}=\frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}\)
119888
A point \(P\) lies on the circle \(x^2+y^2=169\). If \(Q=\) \((5,12)\) and \(R=(-12,5)\), then the angle \(\angle Q P R\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, circle \(\mathrm{x}^2+\mathrm{y}^2=169\) Let point \(\mathrm{p}(\mathrm{x}, \mathrm{y}), \mathrm{q}=(5,12), \mathrm{R}=(-12,5)\) Now, points \(\mathrm{Q}\) and \(\mathrm{R}\) also lies on the circle Slope of line \(\mathrm{OR}=\frac{12}{5}=\mathrm{m}_1\) Slope of line \(\mathrm{OR}=\frac{-5}{12}=\mathrm{m}_2\) We know that, Angle suspended by any chord in circle to circumference is half of angle formed at centre So, \(\quad \angle \mathrm{QOR}=\frac{\pi}{2}\) So, \(\quad \angle \mathrm{QPR}=\frac{1}{2} \angle \mathrm{QOR}=\frac{\pi}{4}\)
WB JEE-2013
Conic Section
119889
If the lengths of the tangents from the point \((1\), 2) to the circles \(x^2+y^2+x+y-4=0\) and \(3 x^2+\) \(3 y^2-x-y-\lambda=0\) are in the ratio \(4: 3\), then the value of \(\lambda\) is
1 \(\frac{21}{2}\)
2 \(\frac{21}{4}\)
3 \(\frac{21}{5}\)
4 \(\frac{21}{11}\)
Explanation:
B Let \(\mathrm{S}_1\) and \(\mathrm{S}_2\) be the two circles given that. \(S_1 =x^2+y^2+x+y-4=0\) \(S_2 =3 x^2+3 y^2-x-y-\lambda=0\) \(=x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0\) Length of a tangent drawn from a point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to a circle \(x^2+y^2+2 g x+2 f y+c=0\) Length of tangent \(L=\sqrt{x_1{ }^2+y_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let \(\mathrm{L}_1\) and \(\mathrm{L}_2\) be the length of tangent drawn from \((1,2)\) \(\text { Given, } \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{4}{3}=\frac{\sqrt{1^2+2^2+1+2-4}}{\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}}\) \(\frac{16}{9}=\frac{4}{4-\frac{\lambda}{3}}\) \(16-\frac{4 \lambda}{3}=9\) \(48-4 \lambda=27\) \(4 \lambda=48-27\) \(4 \lambda=21\) \(\lambda=\frac{21}{4}\)
Jamia Millia Islamia-2010
Conic Section
119891
Suppose the tangents drawn to the circle \(x^2+y^2\) \(-6 x-4 y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P, A\) and \(B\) is
1 \((2,5)\)
2 \((-2,-5)\)
3 \((-2,5)\)
4 \((2,-5)\)
Explanation:
A Given equation of circle. \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-4 \mathrm{y}-11=0\) Equation of \(\mathrm{AB}\) of chord of contact of \(\mathrm{P}\) is using general equation of circle is \(\mathrm{x}_1 \mathrm{x}+\mathrm{y}_1 \mathrm{y}+\mathrm{g}\left(\mathrm{x}_1+\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}_1+\mathrm{y}\right)+\mathrm{c}=0\) we get, at point \((1,8)\) \(\begin{aligned} & 1 x+8 y-3(x+1)-2(y+8)-11=0 \\ & -2 x+6 y-30=0 \\ & x-3 y+15=0 \end{aligned}\) Equation of any circle through \(A B\) is \(x^2+y^2-6 x-4 y-11+\lambda(x-3 y+15)=0\) It will pass through \(P(1,8)\) if \(\begin{aligned} & 1+64-6-32-11+\lambda(1-24+15)=0 \\ & 16-8 \lambda=0 \\ & \lambda=2 \end{aligned}\) By putting the value of \(\lambda\) in equation (i) \(\begin{aligned} & x^2+y^2-6 x-4 y-11+2 x-6 y+30=0 \\ & x^2+y^2-4 x-10 y+19=0 \end{aligned}\) Thus centre of given circle is \((-g,-f)=(2,5)\)
AP EAMCET-06.07.2022
Conic Section
119892
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
1 \(\left(\frac{-3}{2}, 0\right)\)
2 \(\left(\frac{-5}{2}, 2\right)\)
3 \(\left(\frac{-3}{2}, \frac{5}{2}\right)\)
4 \((-4,0)\)
Explanation:
D Since circle touching y-axis at \((0,2)\) is \((x-0)^2+(y-2)^2+\lambda x=0\) passes through \((-1,0)\) \(\therefore 1+4-\lambda=0\) \(\lambda=5\) \(\therefore \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-4 \mathrm{y}+4=0\) Put value of option in equation of circle we get, circle passes through \((-4,0)\)
AP EAMCET-06.07.2022
Conic Section
119893
The slop of the normal to the circle \(x^2+y^2+\) \(\mathbf{2 g x}+\mathbf{2 f y}+\mathbf{c}=\mathbf{0}\) at \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is
1 \(-\left(\frac{x_1+g}{y_1+f}\right)\)
2 \(-\left(\frac{y_1+f}{x_1+g}\right)\)
3 \(\frac{x_1+g}{y_1+f}\)
4 \(\frac{y_1+f}{x_1+g}\)
Explanation:
D Given circle equation \(x^2+y^2+2 g x+2 f y+c=0 \text { at }\left(x_1, y_1\right)\) \(\text { If we know the slope of the tangent we can find the }\) \(\text { equation of tangent at }\left(x_1, y_2\right)\) Tangent and the radius \(\mathrm{CP}\) are perpendicular to each other Slope of tangent \(\times\) slope of \(\mathrm{cp}=-1\) \(\text { slope of tangent }=\frac{-1}{\text { slope of } \mathrm{CP}}\)slope of \(\mathrm{CP}=\frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}\)
119888
A point \(P\) lies on the circle \(x^2+y^2=169\). If \(Q=\) \((5,12)\) and \(R=(-12,5)\), then the angle \(\angle Q P R\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, circle \(\mathrm{x}^2+\mathrm{y}^2=169\) Let point \(\mathrm{p}(\mathrm{x}, \mathrm{y}), \mathrm{q}=(5,12), \mathrm{R}=(-12,5)\) Now, points \(\mathrm{Q}\) and \(\mathrm{R}\) also lies on the circle Slope of line \(\mathrm{OR}=\frac{12}{5}=\mathrm{m}_1\) Slope of line \(\mathrm{OR}=\frac{-5}{12}=\mathrm{m}_2\) We know that, Angle suspended by any chord in circle to circumference is half of angle formed at centre So, \(\quad \angle \mathrm{QOR}=\frac{\pi}{2}\) So, \(\quad \angle \mathrm{QPR}=\frac{1}{2} \angle \mathrm{QOR}=\frac{\pi}{4}\)
WB JEE-2013
Conic Section
119889
If the lengths of the tangents from the point \((1\), 2) to the circles \(x^2+y^2+x+y-4=0\) and \(3 x^2+\) \(3 y^2-x-y-\lambda=0\) are in the ratio \(4: 3\), then the value of \(\lambda\) is
1 \(\frac{21}{2}\)
2 \(\frac{21}{4}\)
3 \(\frac{21}{5}\)
4 \(\frac{21}{11}\)
Explanation:
B Let \(\mathrm{S}_1\) and \(\mathrm{S}_2\) be the two circles given that. \(S_1 =x^2+y^2+x+y-4=0\) \(S_2 =3 x^2+3 y^2-x-y-\lambda=0\) \(=x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0\) Length of a tangent drawn from a point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to a circle \(x^2+y^2+2 g x+2 f y+c=0\) Length of tangent \(L=\sqrt{x_1{ }^2+y_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let \(\mathrm{L}_1\) and \(\mathrm{L}_2\) be the length of tangent drawn from \((1,2)\) \(\text { Given, } \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{4}{3}=\frac{\sqrt{1^2+2^2+1+2-4}}{\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}}\) \(\frac{16}{9}=\frac{4}{4-\frac{\lambda}{3}}\) \(16-\frac{4 \lambda}{3}=9\) \(48-4 \lambda=27\) \(4 \lambda=48-27\) \(4 \lambda=21\) \(\lambda=\frac{21}{4}\)
Jamia Millia Islamia-2010
Conic Section
119891
Suppose the tangents drawn to the circle \(x^2+y^2\) \(-6 x-4 y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P, A\) and \(B\) is
1 \((2,5)\)
2 \((-2,-5)\)
3 \((-2,5)\)
4 \((2,-5)\)
Explanation:
A Given equation of circle. \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-4 \mathrm{y}-11=0\) Equation of \(\mathrm{AB}\) of chord of contact of \(\mathrm{P}\) is using general equation of circle is \(\mathrm{x}_1 \mathrm{x}+\mathrm{y}_1 \mathrm{y}+\mathrm{g}\left(\mathrm{x}_1+\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}_1+\mathrm{y}\right)+\mathrm{c}=0\) we get, at point \((1,8)\) \(\begin{aligned} & 1 x+8 y-3(x+1)-2(y+8)-11=0 \\ & -2 x+6 y-30=0 \\ & x-3 y+15=0 \end{aligned}\) Equation of any circle through \(A B\) is \(x^2+y^2-6 x-4 y-11+\lambda(x-3 y+15)=0\) It will pass through \(P(1,8)\) if \(\begin{aligned} & 1+64-6-32-11+\lambda(1-24+15)=0 \\ & 16-8 \lambda=0 \\ & \lambda=2 \end{aligned}\) By putting the value of \(\lambda\) in equation (i) \(\begin{aligned} & x^2+y^2-6 x-4 y-11+2 x-6 y+30=0 \\ & x^2+y^2-4 x-10 y+19=0 \end{aligned}\) Thus centre of given circle is \((-g,-f)=(2,5)\)
AP EAMCET-06.07.2022
Conic Section
119892
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
1 \(\left(\frac{-3}{2}, 0\right)\)
2 \(\left(\frac{-5}{2}, 2\right)\)
3 \(\left(\frac{-3}{2}, \frac{5}{2}\right)\)
4 \((-4,0)\)
Explanation:
D Since circle touching y-axis at \((0,2)\) is \((x-0)^2+(y-2)^2+\lambda x=0\) passes through \((-1,0)\) \(\therefore 1+4-\lambda=0\) \(\lambda=5\) \(\therefore \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-4 \mathrm{y}+4=0\) Put value of option in equation of circle we get, circle passes through \((-4,0)\)
AP EAMCET-06.07.2022
Conic Section
119893
The slop of the normal to the circle \(x^2+y^2+\) \(\mathbf{2 g x}+\mathbf{2 f y}+\mathbf{c}=\mathbf{0}\) at \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is
1 \(-\left(\frac{x_1+g}{y_1+f}\right)\)
2 \(-\left(\frac{y_1+f}{x_1+g}\right)\)
3 \(\frac{x_1+g}{y_1+f}\)
4 \(\frac{y_1+f}{x_1+g}\)
Explanation:
D Given circle equation \(x^2+y^2+2 g x+2 f y+c=0 \text { at }\left(x_1, y_1\right)\) \(\text { If we know the slope of the tangent we can find the }\) \(\text { equation of tangent at }\left(x_1, y_2\right)\) Tangent and the radius \(\mathrm{CP}\) are perpendicular to each other Slope of tangent \(\times\) slope of \(\mathrm{cp}=-1\) \(\text { slope of tangent }=\frac{-1}{\text { slope of } \mathrm{CP}}\)slope of \(\mathrm{CP}=\frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}\)
119888
A point \(P\) lies on the circle \(x^2+y^2=169\). If \(Q=\) \((5,12)\) and \(R=(-12,5)\), then the angle \(\angle Q P R\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, circle \(\mathrm{x}^2+\mathrm{y}^2=169\) Let point \(\mathrm{p}(\mathrm{x}, \mathrm{y}), \mathrm{q}=(5,12), \mathrm{R}=(-12,5)\) Now, points \(\mathrm{Q}\) and \(\mathrm{R}\) also lies on the circle Slope of line \(\mathrm{OR}=\frac{12}{5}=\mathrm{m}_1\) Slope of line \(\mathrm{OR}=\frac{-5}{12}=\mathrm{m}_2\) We know that, Angle suspended by any chord in circle to circumference is half of angle formed at centre So, \(\quad \angle \mathrm{QOR}=\frac{\pi}{2}\) So, \(\quad \angle \mathrm{QPR}=\frac{1}{2} \angle \mathrm{QOR}=\frac{\pi}{4}\)
WB JEE-2013
Conic Section
119889
If the lengths of the tangents from the point \((1\), 2) to the circles \(x^2+y^2+x+y-4=0\) and \(3 x^2+\) \(3 y^2-x-y-\lambda=0\) are in the ratio \(4: 3\), then the value of \(\lambda\) is
1 \(\frac{21}{2}\)
2 \(\frac{21}{4}\)
3 \(\frac{21}{5}\)
4 \(\frac{21}{11}\)
Explanation:
B Let \(\mathrm{S}_1\) and \(\mathrm{S}_2\) be the two circles given that. \(S_1 =x^2+y^2+x+y-4=0\) \(S_2 =3 x^2+3 y^2-x-y-\lambda=0\) \(=x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0\) Length of a tangent drawn from a point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to a circle \(x^2+y^2+2 g x+2 f y+c=0\) Length of tangent \(L=\sqrt{x_1{ }^2+y_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let \(\mathrm{L}_1\) and \(\mathrm{L}_2\) be the length of tangent drawn from \((1,2)\) \(\text { Given, } \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{4}{3}=\frac{\sqrt{1^2+2^2+1+2-4}}{\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}}\) \(\frac{16}{9}=\frac{4}{4-\frac{\lambda}{3}}\) \(16-\frac{4 \lambda}{3}=9\) \(48-4 \lambda=27\) \(4 \lambda=48-27\) \(4 \lambda=21\) \(\lambda=\frac{21}{4}\)
Jamia Millia Islamia-2010
Conic Section
119891
Suppose the tangents drawn to the circle \(x^2+y^2\) \(-6 x-4 y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P, A\) and \(B\) is
1 \((2,5)\)
2 \((-2,-5)\)
3 \((-2,5)\)
4 \((2,-5)\)
Explanation:
A Given equation of circle. \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-4 \mathrm{y}-11=0\) Equation of \(\mathrm{AB}\) of chord of contact of \(\mathrm{P}\) is using general equation of circle is \(\mathrm{x}_1 \mathrm{x}+\mathrm{y}_1 \mathrm{y}+\mathrm{g}\left(\mathrm{x}_1+\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}_1+\mathrm{y}\right)+\mathrm{c}=0\) we get, at point \((1,8)\) \(\begin{aligned} & 1 x+8 y-3(x+1)-2(y+8)-11=0 \\ & -2 x+6 y-30=0 \\ & x-3 y+15=0 \end{aligned}\) Equation of any circle through \(A B\) is \(x^2+y^2-6 x-4 y-11+\lambda(x-3 y+15)=0\) It will pass through \(P(1,8)\) if \(\begin{aligned} & 1+64-6-32-11+\lambda(1-24+15)=0 \\ & 16-8 \lambda=0 \\ & \lambda=2 \end{aligned}\) By putting the value of \(\lambda\) in equation (i) \(\begin{aligned} & x^2+y^2-6 x-4 y-11+2 x-6 y+30=0 \\ & x^2+y^2-4 x-10 y+19=0 \end{aligned}\) Thus centre of given circle is \((-g,-f)=(2,5)\)
AP EAMCET-06.07.2022
Conic Section
119892
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
1 \(\left(\frac{-3}{2}, 0\right)\)
2 \(\left(\frac{-5}{2}, 2\right)\)
3 \(\left(\frac{-3}{2}, \frac{5}{2}\right)\)
4 \((-4,0)\)
Explanation:
D Since circle touching y-axis at \((0,2)\) is \((x-0)^2+(y-2)^2+\lambda x=0\) passes through \((-1,0)\) \(\therefore 1+4-\lambda=0\) \(\lambda=5\) \(\therefore \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-4 \mathrm{y}+4=0\) Put value of option in equation of circle we get, circle passes through \((-4,0)\)
AP EAMCET-06.07.2022
Conic Section
119893
The slop of the normal to the circle \(x^2+y^2+\) \(\mathbf{2 g x}+\mathbf{2 f y}+\mathbf{c}=\mathbf{0}\) at \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is
1 \(-\left(\frac{x_1+g}{y_1+f}\right)\)
2 \(-\left(\frac{y_1+f}{x_1+g}\right)\)
3 \(\frac{x_1+g}{y_1+f}\)
4 \(\frac{y_1+f}{x_1+g}\)
Explanation:
D Given circle equation \(x^2+y^2+2 g x+2 f y+c=0 \text { at }\left(x_1, y_1\right)\) \(\text { If we know the slope of the tangent we can find the }\) \(\text { equation of tangent at }\left(x_1, y_2\right)\) Tangent and the radius \(\mathrm{CP}\) are perpendicular to each other Slope of tangent \(\times\) slope of \(\mathrm{cp}=-1\) \(\text { slope of tangent }=\frac{-1}{\text { slope of } \mathrm{CP}}\)slope of \(\mathrm{CP}=\frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}\)
