119705
Let \(A=(0,4)\) and \(B=(2 \cos \theta, 2 \sin \theta)\), for some 0 \(\lt \theta\lt \frac{\pi}{2}\). Let \(P\) divide the line segment \(A B\) in the ratio \(2: 3\) internally. The locus of \(P\) is
1 Circle
2 Ellipse
3 Parabola
4 Hyperbola
Explanation:
A It is given that, By using section formula : \(\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}\) \(\mathrm{x}=\frac{4 \cos \theta+0}{5}\) \(\mathrm{y}=\frac{4 \sin \theta+12}{5}\) Solve the equation (i) for \(\cos \theta\) as follows \(\cos \theta=\frac{5 \mathrm{x}}{4}\) Solve the equation (ii) for \(\sin \theta\) as follows \(\sin \theta=\frac{5 y-12}{4}\) Square and add equation (iii) and (iv) - \(\cos ^2 \theta+\sin ^2 \theta=1\) \(\left(\frac{5 x}{4}\right)^2+\left(\frac{5 y-12}{4}\right)^2=1\) \(\frac{25 x^2}{16}+\frac{(5 y-12)^2}{16}=1\) \(25 x^2+(5 y-12)^2=16\)This represents the equation of circle.
TS EAMCET-09.09.2020
Conic Section
119706
The equation of the circle touching the line \(2 x+3 y+1=0\) at the point \((1,-1)\) and orthogonal to the circle which has the line segment having end points \((0,-1)\) and \((-2,3)\) as diameter, is
1 \(x^2+y^2-10 x+5 y+1=0\)
2 \(x^2+y^2+5 x-10 y-1=0\)
3 \(2 x^2+2 y^2+10 x-5 y-1=0\)
4 \(2 x^2+2 y^2-10 x-5 y+1=0\)
Explanation:
D We are given that line \(2 x+3 y+1=0\) touches a circle \(S=0\) at \((1,-1)\) The equation of the circle- \((x-1)^2+(y+1)^2+\lambda(2 x+3 y+1)=0\) \(x^2+y^2+2 x(\lambda-1)+y(3 \lambda+2)+(\lambda+2)=0\) \(\text { i.e } x(x+2)+(y+1)(y-3)=0\) \(x^2+y^2+2 x-2 y-3=0\) \(2\left(g_1 g_2+f_1 \mathrm{f}_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\) \(2 x^2+2 y^2-10 x-5 y+1=0\) But given that this circle is orthogonal to the circle, the extremities of whose diameter are \((0,-1)\) and \((-2,3)\) Applying the condition for orthogonalality - \(2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\)Putting the value of \(\lambda\) in equation (i) the required circle equation is-
TS EAMCET-04.05.2019
Conic Section
119707
If one of the diameters of the circle \(x^2+y^2-2 x\) \(-6 y+6=0\) is a chord to the circle with centre \((2,1)\), then the radius of the bigger circle is
1 6
2 4
3 2
4 3
Explanation:
D Given, circle equation is- \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-6 \mathrm{y}+6=0\) Centre \((1,3), r=\sqrt{1+9-6}=2\) \(\mathrm{OA}=\sqrt{(2-1)^2(1-3)^2}\) \(\mathrm{OA}=\sqrt{5}\) \(\because \mathrm{AB}=\mathrm{r}=2, \mathrm{R}^2=\mathrm{r}^2+\mathrm{OA}^2\) \(\mathrm{R}^2=2^2+(\sqrt{5})^2\) \(=4+5\) \(\mathrm{R}^2=9\) \(\mathrm{R}=\sqrt{9}=3\)
TS EAMCET-03.05.2019
Conic Section
119708
If the points \((2,3)\) and \((K,-2)\) are conjugate with respect to the circle \(x^2+y^2-2 x+4 y-2=\) 0 then \(\mathrm{K}=\)
1 8
2 6
3 4
4 3
Explanation:
A Equation of circle \(\quad \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-2=0\) \(\text { Polar of point }(2,3) \text { is }\) \(\mathrm{S}_1=0\) \(2 \mathrm{x}+3 \mathrm{y}-2\left(\frac{\mathrm{x}+2}{2}\right)+4\left(\frac{\mathrm{y}+3}{2}\right)-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}-2+2 \mathrm{y}+6-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}+2 \mathrm{y}+2=0\) \(\Rightarrow \mathrm{x}+5 \mathrm{y}+2=0\) \(\text { Given that, the points }(2,3) \text { and }(\mathrm{K},-2) \text { are conjugate }\) \(\text { Therefore, polar of point }(\mathrm{K},-2)\)Given that, the points \((2,3)\) and \((\mathrm{K},-2)\) are conjugate Therefore, polar of point \((\mathrm{K},-2)\) \(\therefore \quad \mathrm{K}-10+2=0\) \(\Rightarrow \mathrm{K}-8=0\) \(\Rightarrow \mathrm{K}=8\)Hence, the value of \(\mathrm{K}\) is 8 .
119705
Let \(A=(0,4)\) and \(B=(2 \cos \theta, 2 \sin \theta)\), for some 0 \(\lt \theta\lt \frac{\pi}{2}\). Let \(P\) divide the line segment \(A B\) in the ratio \(2: 3\) internally. The locus of \(P\) is
1 Circle
2 Ellipse
3 Parabola
4 Hyperbola
Explanation:
A It is given that, By using section formula : \(\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}\) \(\mathrm{x}=\frac{4 \cos \theta+0}{5}\) \(\mathrm{y}=\frac{4 \sin \theta+12}{5}\) Solve the equation (i) for \(\cos \theta\) as follows \(\cos \theta=\frac{5 \mathrm{x}}{4}\) Solve the equation (ii) for \(\sin \theta\) as follows \(\sin \theta=\frac{5 y-12}{4}\) Square and add equation (iii) and (iv) - \(\cos ^2 \theta+\sin ^2 \theta=1\) \(\left(\frac{5 x}{4}\right)^2+\left(\frac{5 y-12}{4}\right)^2=1\) \(\frac{25 x^2}{16}+\frac{(5 y-12)^2}{16}=1\) \(25 x^2+(5 y-12)^2=16\)This represents the equation of circle.
TS EAMCET-09.09.2020
Conic Section
119706
The equation of the circle touching the line \(2 x+3 y+1=0\) at the point \((1,-1)\) and orthogonal to the circle which has the line segment having end points \((0,-1)\) and \((-2,3)\) as diameter, is
1 \(x^2+y^2-10 x+5 y+1=0\)
2 \(x^2+y^2+5 x-10 y-1=0\)
3 \(2 x^2+2 y^2+10 x-5 y-1=0\)
4 \(2 x^2+2 y^2-10 x-5 y+1=0\)
Explanation:
D We are given that line \(2 x+3 y+1=0\) touches a circle \(S=0\) at \((1,-1)\) The equation of the circle- \((x-1)^2+(y+1)^2+\lambda(2 x+3 y+1)=0\) \(x^2+y^2+2 x(\lambda-1)+y(3 \lambda+2)+(\lambda+2)=0\) \(\text { i.e } x(x+2)+(y+1)(y-3)=0\) \(x^2+y^2+2 x-2 y-3=0\) \(2\left(g_1 g_2+f_1 \mathrm{f}_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\) \(2 x^2+2 y^2-10 x-5 y+1=0\) But given that this circle is orthogonal to the circle, the extremities of whose diameter are \((0,-1)\) and \((-2,3)\) Applying the condition for orthogonalality - \(2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\)Putting the value of \(\lambda\) in equation (i) the required circle equation is-
TS EAMCET-04.05.2019
Conic Section
119707
If one of the diameters of the circle \(x^2+y^2-2 x\) \(-6 y+6=0\) is a chord to the circle with centre \((2,1)\), then the radius of the bigger circle is
1 6
2 4
3 2
4 3
Explanation:
D Given, circle equation is- \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-6 \mathrm{y}+6=0\) Centre \((1,3), r=\sqrt{1+9-6}=2\) \(\mathrm{OA}=\sqrt{(2-1)^2(1-3)^2}\) \(\mathrm{OA}=\sqrt{5}\) \(\because \mathrm{AB}=\mathrm{r}=2, \mathrm{R}^2=\mathrm{r}^2+\mathrm{OA}^2\) \(\mathrm{R}^2=2^2+(\sqrt{5})^2\) \(=4+5\) \(\mathrm{R}^2=9\) \(\mathrm{R}=\sqrt{9}=3\)
TS EAMCET-03.05.2019
Conic Section
119708
If the points \((2,3)\) and \((K,-2)\) are conjugate with respect to the circle \(x^2+y^2-2 x+4 y-2=\) 0 then \(\mathrm{K}=\)
1 8
2 6
3 4
4 3
Explanation:
A Equation of circle \(\quad \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-2=0\) \(\text { Polar of point }(2,3) \text { is }\) \(\mathrm{S}_1=0\) \(2 \mathrm{x}+3 \mathrm{y}-2\left(\frac{\mathrm{x}+2}{2}\right)+4\left(\frac{\mathrm{y}+3}{2}\right)-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}-2+2 \mathrm{y}+6-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}+2 \mathrm{y}+2=0\) \(\Rightarrow \mathrm{x}+5 \mathrm{y}+2=0\) \(\text { Given that, the points }(2,3) \text { and }(\mathrm{K},-2) \text { are conjugate }\) \(\text { Therefore, polar of point }(\mathrm{K},-2)\)Given that, the points \((2,3)\) and \((\mathrm{K},-2)\) are conjugate Therefore, polar of point \((\mathrm{K},-2)\) \(\therefore \quad \mathrm{K}-10+2=0\) \(\Rightarrow \mathrm{K}-8=0\) \(\Rightarrow \mathrm{K}=8\)Hence, the value of \(\mathrm{K}\) is 8 .
119705
Let \(A=(0,4)\) and \(B=(2 \cos \theta, 2 \sin \theta)\), for some 0 \(\lt \theta\lt \frac{\pi}{2}\). Let \(P\) divide the line segment \(A B\) in the ratio \(2: 3\) internally. The locus of \(P\) is
1 Circle
2 Ellipse
3 Parabola
4 Hyperbola
Explanation:
A It is given that, By using section formula : \(\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}\) \(\mathrm{x}=\frac{4 \cos \theta+0}{5}\) \(\mathrm{y}=\frac{4 \sin \theta+12}{5}\) Solve the equation (i) for \(\cos \theta\) as follows \(\cos \theta=\frac{5 \mathrm{x}}{4}\) Solve the equation (ii) for \(\sin \theta\) as follows \(\sin \theta=\frac{5 y-12}{4}\) Square and add equation (iii) and (iv) - \(\cos ^2 \theta+\sin ^2 \theta=1\) \(\left(\frac{5 x}{4}\right)^2+\left(\frac{5 y-12}{4}\right)^2=1\) \(\frac{25 x^2}{16}+\frac{(5 y-12)^2}{16}=1\) \(25 x^2+(5 y-12)^2=16\)This represents the equation of circle.
TS EAMCET-09.09.2020
Conic Section
119706
The equation of the circle touching the line \(2 x+3 y+1=0\) at the point \((1,-1)\) and orthogonal to the circle which has the line segment having end points \((0,-1)\) and \((-2,3)\) as diameter, is
1 \(x^2+y^2-10 x+5 y+1=0\)
2 \(x^2+y^2+5 x-10 y-1=0\)
3 \(2 x^2+2 y^2+10 x-5 y-1=0\)
4 \(2 x^2+2 y^2-10 x-5 y+1=0\)
Explanation:
D We are given that line \(2 x+3 y+1=0\) touches a circle \(S=0\) at \((1,-1)\) The equation of the circle- \((x-1)^2+(y+1)^2+\lambda(2 x+3 y+1)=0\) \(x^2+y^2+2 x(\lambda-1)+y(3 \lambda+2)+(\lambda+2)=0\) \(\text { i.e } x(x+2)+(y+1)(y-3)=0\) \(x^2+y^2+2 x-2 y-3=0\) \(2\left(g_1 g_2+f_1 \mathrm{f}_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\) \(2 x^2+2 y^2-10 x-5 y+1=0\) But given that this circle is orthogonal to the circle, the extremities of whose diameter are \((0,-1)\) and \((-2,3)\) Applying the condition for orthogonalality - \(2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\)Putting the value of \(\lambda\) in equation (i) the required circle equation is-
TS EAMCET-04.05.2019
Conic Section
119707
If one of the diameters of the circle \(x^2+y^2-2 x\) \(-6 y+6=0\) is a chord to the circle with centre \((2,1)\), then the radius of the bigger circle is
1 6
2 4
3 2
4 3
Explanation:
D Given, circle equation is- \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-6 \mathrm{y}+6=0\) Centre \((1,3), r=\sqrt{1+9-6}=2\) \(\mathrm{OA}=\sqrt{(2-1)^2(1-3)^2}\) \(\mathrm{OA}=\sqrt{5}\) \(\because \mathrm{AB}=\mathrm{r}=2, \mathrm{R}^2=\mathrm{r}^2+\mathrm{OA}^2\) \(\mathrm{R}^2=2^2+(\sqrt{5})^2\) \(=4+5\) \(\mathrm{R}^2=9\) \(\mathrm{R}=\sqrt{9}=3\)
TS EAMCET-03.05.2019
Conic Section
119708
If the points \((2,3)\) and \((K,-2)\) are conjugate with respect to the circle \(x^2+y^2-2 x+4 y-2=\) 0 then \(\mathrm{K}=\)
1 8
2 6
3 4
4 3
Explanation:
A Equation of circle \(\quad \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-2=0\) \(\text { Polar of point }(2,3) \text { is }\) \(\mathrm{S}_1=0\) \(2 \mathrm{x}+3 \mathrm{y}-2\left(\frac{\mathrm{x}+2}{2}\right)+4\left(\frac{\mathrm{y}+3}{2}\right)-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}-2+2 \mathrm{y}+6-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}+2 \mathrm{y}+2=0\) \(\Rightarrow \mathrm{x}+5 \mathrm{y}+2=0\) \(\text { Given that, the points }(2,3) \text { and }(\mathrm{K},-2) \text { are conjugate }\) \(\text { Therefore, polar of point }(\mathrm{K},-2)\)Given that, the points \((2,3)\) and \((\mathrm{K},-2)\) are conjugate Therefore, polar of point \((\mathrm{K},-2)\) \(\therefore \quad \mathrm{K}-10+2=0\) \(\Rightarrow \mathrm{K}-8=0\) \(\Rightarrow \mathrm{K}=8\)Hence, the value of \(\mathrm{K}\) is 8 .
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119705
Let \(A=(0,4)\) and \(B=(2 \cos \theta, 2 \sin \theta)\), for some 0 \(\lt \theta\lt \frac{\pi}{2}\). Let \(P\) divide the line segment \(A B\) in the ratio \(2: 3\) internally. The locus of \(P\) is
1 Circle
2 Ellipse
3 Parabola
4 Hyperbola
Explanation:
A It is given that, By using section formula : \(\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}\) \(\mathrm{x}=\frac{4 \cos \theta+0}{5}\) \(\mathrm{y}=\frac{4 \sin \theta+12}{5}\) Solve the equation (i) for \(\cos \theta\) as follows \(\cos \theta=\frac{5 \mathrm{x}}{4}\) Solve the equation (ii) for \(\sin \theta\) as follows \(\sin \theta=\frac{5 y-12}{4}\) Square and add equation (iii) and (iv) - \(\cos ^2 \theta+\sin ^2 \theta=1\) \(\left(\frac{5 x}{4}\right)^2+\left(\frac{5 y-12}{4}\right)^2=1\) \(\frac{25 x^2}{16}+\frac{(5 y-12)^2}{16}=1\) \(25 x^2+(5 y-12)^2=16\)This represents the equation of circle.
TS EAMCET-09.09.2020
Conic Section
119706
The equation of the circle touching the line \(2 x+3 y+1=0\) at the point \((1,-1)\) and orthogonal to the circle which has the line segment having end points \((0,-1)\) and \((-2,3)\) as diameter, is
1 \(x^2+y^2-10 x+5 y+1=0\)
2 \(x^2+y^2+5 x-10 y-1=0\)
3 \(2 x^2+2 y^2+10 x-5 y-1=0\)
4 \(2 x^2+2 y^2-10 x-5 y+1=0\)
Explanation:
D We are given that line \(2 x+3 y+1=0\) touches a circle \(S=0\) at \((1,-1)\) The equation of the circle- \((x-1)^2+(y+1)^2+\lambda(2 x+3 y+1)=0\) \(x^2+y^2+2 x(\lambda-1)+y(3 \lambda+2)+(\lambda+2)=0\) \(\text { i.e } x(x+2)+(y+1)(y-3)=0\) \(x^2+y^2+2 x-2 y-3=0\) \(2\left(g_1 g_2+f_1 \mathrm{f}_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\) \(2 x^2+2 y^2-10 x-5 y+1=0\) But given that this circle is orthogonal to the circle, the extremities of whose diameter are \((0,-1)\) and \((-2,3)\) Applying the condition for orthogonalality - \(2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2\) \(2(\lambda-1) \cdot 1+2\left(1+\frac{3 \lambda}{2}\right)(-1)=2+\lambda-3\) \(-\lambda-4=\lambda-1\) \(2 \lambda=-3\) \(\lambda=\frac{-3}{2}\)Putting the value of \(\lambda\) in equation (i) the required circle equation is-
TS EAMCET-04.05.2019
Conic Section
119707
If one of the diameters of the circle \(x^2+y^2-2 x\) \(-6 y+6=0\) is a chord to the circle with centre \((2,1)\), then the radius of the bigger circle is
1 6
2 4
3 2
4 3
Explanation:
D Given, circle equation is- \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-6 \mathrm{y}+6=0\) Centre \((1,3), r=\sqrt{1+9-6}=2\) \(\mathrm{OA}=\sqrt{(2-1)^2(1-3)^2}\) \(\mathrm{OA}=\sqrt{5}\) \(\because \mathrm{AB}=\mathrm{r}=2, \mathrm{R}^2=\mathrm{r}^2+\mathrm{OA}^2\) \(\mathrm{R}^2=2^2+(\sqrt{5})^2\) \(=4+5\) \(\mathrm{R}^2=9\) \(\mathrm{R}=\sqrt{9}=3\)
TS EAMCET-03.05.2019
Conic Section
119708
If the points \((2,3)\) and \((K,-2)\) are conjugate with respect to the circle \(x^2+y^2-2 x+4 y-2=\) 0 then \(\mathrm{K}=\)
1 8
2 6
3 4
4 3
Explanation:
A Equation of circle \(\quad \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-2=0\) \(\text { Polar of point }(2,3) \text { is }\) \(\mathrm{S}_1=0\) \(2 \mathrm{x}+3 \mathrm{y}-2\left(\frac{\mathrm{x}+2}{2}\right)+4\left(\frac{\mathrm{y}+3}{2}\right)-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}-2+2 \mathrm{y}+6-2=0\) \(\quad \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}+2 \mathrm{y}+2=0\) \(\Rightarrow \mathrm{x}+5 \mathrm{y}+2=0\) \(\text { Given that, the points }(2,3) \text { and }(\mathrm{K},-2) \text { are conjugate }\) \(\text { Therefore, polar of point }(\mathrm{K},-2)\)Given that, the points \((2,3)\) and \((\mathrm{K},-2)\) are conjugate Therefore, polar of point \((\mathrm{K},-2)\) \(\therefore \quad \mathrm{K}-10+2=0\) \(\Rightarrow \mathrm{K}-8=0\) \(\Rightarrow \mathrm{K}=8\)Hence, the value of \(\mathrm{K}\) is 8 .
