120051
If the area of the circle \(7 x^2+7 y^2-7 x+14 y+k\) \(=0\) is \(12 \pi\) sq. units, then the value of \(k\) is
1 \(\frac{-43}{4}\)
2 \(\frac{-301}{4}\)
3 -16
4 \(\pm 4\)
Explanation:
B Equation of circle \(7 x^2+7 y^2-7 x+14 y+k=0\) can be written as, \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+2 \mathrm{y}+\frac{\mathrm{k}}{7}=0\) \(\therefore\) Centre of circle is \(\left(\frac{1}{2},-1\right)\) Radius of circle, \(r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}\) \(=\sqrt{\frac{5}{4}-\frac{\mathrm{k}}{7}} \Rightarrow \mathrm{r}^2=\frac{5}{4}-\frac{\mathrm{k}}{7}\) Given that area of circle \(=12 \pi\) sq. units \(\Rightarrow \pi \mathrm{r}^2=\pi\left(\frac{5}{4}-\frac{\mathrm{k}}{7}\right)=12 \pi\) \(\frac{5}{4}-\frac{\mathrm{k}}{7}=12\) \(\Rightarrow \frac{5}{4}-12=\frac{\mathrm{k}}{7} \Rightarrow \frac{-43}{4}=\frac{\mathrm{k}}{7}\) \(\Rightarrow \mathrm{k}=\frac{-301}{4}\)
COMEDK-2012
Conic Section
120052
If \((-3,2)\) lies on the circle \(x^2+y^2+2 g x+2 f y+c\) \(=0\) which is concentric with the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(6 x+8 y-5=0\), then \(C\) is equal to
1 11
2 -11
3 24
4 100
Explanation:
B If the circle \(x^2+y^2+2 g x+2 f y+c=0\) is concentric with \(x^2+y^2+6 x+8 y-5=0\) then (i) can be written as \(x^2+y^2+6 x+8 y+C=0\) and since it passes through \((-3,2)\), we get \((-3)^2+(2)^2+(6)(-3)+8 \times 2+\mathrm{C}=0\) \(9+4-18+16+\mathrm{C}=0\) \(\mathrm{C}=-11\)
BCECE-2013
Conic Section
120053
The locus of the centers of the circle that are passing through the intersection of the circles \(x^2+y^2=1\) and \(x^2+y^2-2 x+y=0\) is
1 A line whose equation is \(x+2 y=0\)
2 A circle
3 A parabola
4 A line whose equation is \(2 x-y=0\)
Explanation:
A : Let \(S_1=x^2+y^2=1\) \(S_2=x^2+y^2-2 x+y=0\) The equation of a family of circles passing through the intersection of given circle is:- \(S_1+\lambda S_2=0\) \(x^2+y^2-\frac{2 \lambda}{1+\lambda} \mathrm{x}+\frac{\lambda}{1+\lambda} \mathrm{y}-\frac{1}{1+\lambda}=0=0\) Let, \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) be its centre. Then, \(\mathrm{h}=\frac{\lambda}{1+\lambda}\) and \(\mathrm{k}=\) \(-\left(\frac{\lambda}{2(1+\lambda)}\right)\) \(\Rightarrow \mathrm{h}=-2 \mathrm{k}\) Hence, the locus of \((\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}=-2 \mathrm{y}\) \(x+2 y=0\)
APEAPCET-20.08.2021
Conic Section
120054
The point which has the same power with respect to each of the circles \(x^2+y^2-8 x+40=0, x^2+y^2-5 x+16=0\) and \(x^2+y^2-8 x+16 y+160=0\) is
1 \(\left(-8, \frac{-15}{2}\right)\)
2 \(\left(8, \frac{-15}{2}\right)\)
3 \(\left(8, \frac{15}{2}\right)\)
4 \(\left(-8, \frac{15}{2}\right)\)
Explanation:
B Given, \(S_1: x^2+y^2-8 x+40=0\) \(S_2: x^2+y^2-5 x+16=0\) \(S_3: x^2+y^2-8 x+16 y+160=0\) So, \(\mathrm{S}_2-\mathrm{S}_1=0\) \(3 \mathrm{x}-24=0\) \(\mathrm{x}=8\) and, \(S_3-S_1=0\) \(16 y+120=0\) \(4 y+30=0\) \(y=\frac{-15}{2}\) So, Point is \(\left(8, \frac{-15}{2}\right)\) Hence, required point is \(\left(8, \frac{-15}{2}\right)\).
AP EAMCET-19.08.2021
Conic Section
120055
If the chord of contact of tangents from a point on the circle \(x^2+y^2=r_1^2\) to the circle \(x^2+y^2=r_2{ }^2\) touches the circle \(x^2+y^2=r_3{ }^2\) then \(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3\) are in
1 \(\mathrm{AP}\)
2 \(\mathrm{HP}\)
3 GP
4 AGP
Explanation:
C : Assume, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be any point on the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) then, \(\mathrm{x}_1^2+\mathrm{y}_1{ }^2=\mathrm{r}_1{ }^2\) Equation of chord of contact of tangents from \(\left(x_1, y_1\right)\) to the circle \(x^2+y^2=r_2^2\) is \(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{r}_2{ }^2\) (ii) touches the circle \(x^2+y^2=r_3{ }^2\) \(\therefore\) The length of perpendicular from centre \((0,0)\) on (ii) is \(=\) radius \(=r_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{r}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \mathrm{r}_2^2=\mathrm{r}_1 \mathrm{r}_3\)Hence, \(\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3\) are in G.P.
120051
If the area of the circle \(7 x^2+7 y^2-7 x+14 y+k\) \(=0\) is \(12 \pi\) sq. units, then the value of \(k\) is
1 \(\frac{-43}{4}\)
2 \(\frac{-301}{4}\)
3 -16
4 \(\pm 4\)
Explanation:
B Equation of circle \(7 x^2+7 y^2-7 x+14 y+k=0\) can be written as, \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+2 \mathrm{y}+\frac{\mathrm{k}}{7}=0\) \(\therefore\) Centre of circle is \(\left(\frac{1}{2},-1\right)\) Radius of circle, \(r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}\) \(=\sqrt{\frac{5}{4}-\frac{\mathrm{k}}{7}} \Rightarrow \mathrm{r}^2=\frac{5}{4}-\frac{\mathrm{k}}{7}\) Given that area of circle \(=12 \pi\) sq. units \(\Rightarrow \pi \mathrm{r}^2=\pi\left(\frac{5}{4}-\frac{\mathrm{k}}{7}\right)=12 \pi\) \(\frac{5}{4}-\frac{\mathrm{k}}{7}=12\) \(\Rightarrow \frac{5}{4}-12=\frac{\mathrm{k}}{7} \Rightarrow \frac{-43}{4}=\frac{\mathrm{k}}{7}\) \(\Rightarrow \mathrm{k}=\frac{-301}{4}\)
COMEDK-2012
Conic Section
120052
If \((-3,2)\) lies on the circle \(x^2+y^2+2 g x+2 f y+c\) \(=0\) which is concentric with the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(6 x+8 y-5=0\), then \(C\) is equal to
1 11
2 -11
3 24
4 100
Explanation:
B If the circle \(x^2+y^2+2 g x+2 f y+c=0\) is concentric with \(x^2+y^2+6 x+8 y-5=0\) then (i) can be written as \(x^2+y^2+6 x+8 y+C=0\) and since it passes through \((-3,2)\), we get \((-3)^2+(2)^2+(6)(-3)+8 \times 2+\mathrm{C}=0\) \(9+4-18+16+\mathrm{C}=0\) \(\mathrm{C}=-11\)
BCECE-2013
Conic Section
120053
The locus of the centers of the circle that are passing through the intersection of the circles \(x^2+y^2=1\) and \(x^2+y^2-2 x+y=0\) is
1 A line whose equation is \(x+2 y=0\)
2 A circle
3 A parabola
4 A line whose equation is \(2 x-y=0\)
Explanation:
A : Let \(S_1=x^2+y^2=1\) \(S_2=x^2+y^2-2 x+y=0\) The equation of a family of circles passing through the intersection of given circle is:- \(S_1+\lambda S_2=0\) \(x^2+y^2-\frac{2 \lambda}{1+\lambda} \mathrm{x}+\frac{\lambda}{1+\lambda} \mathrm{y}-\frac{1}{1+\lambda}=0=0\) Let, \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) be its centre. Then, \(\mathrm{h}=\frac{\lambda}{1+\lambda}\) and \(\mathrm{k}=\) \(-\left(\frac{\lambda}{2(1+\lambda)}\right)\) \(\Rightarrow \mathrm{h}=-2 \mathrm{k}\) Hence, the locus of \((\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}=-2 \mathrm{y}\) \(x+2 y=0\)
APEAPCET-20.08.2021
Conic Section
120054
The point which has the same power with respect to each of the circles \(x^2+y^2-8 x+40=0, x^2+y^2-5 x+16=0\) and \(x^2+y^2-8 x+16 y+160=0\) is
1 \(\left(-8, \frac{-15}{2}\right)\)
2 \(\left(8, \frac{-15}{2}\right)\)
3 \(\left(8, \frac{15}{2}\right)\)
4 \(\left(-8, \frac{15}{2}\right)\)
Explanation:
B Given, \(S_1: x^2+y^2-8 x+40=0\) \(S_2: x^2+y^2-5 x+16=0\) \(S_3: x^2+y^2-8 x+16 y+160=0\) So, \(\mathrm{S}_2-\mathrm{S}_1=0\) \(3 \mathrm{x}-24=0\) \(\mathrm{x}=8\) and, \(S_3-S_1=0\) \(16 y+120=0\) \(4 y+30=0\) \(y=\frac{-15}{2}\) So, Point is \(\left(8, \frac{-15}{2}\right)\) Hence, required point is \(\left(8, \frac{-15}{2}\right)\).
AP EAMCET-19.08.2021
Conic Section
120055
If the chord of contact of tangents from a point on the circle \(x^2+y^2=r_1^2\) to the circle \(x^2+y^2=r_2{ }^2\) touches the circle \(x^2+y^2=r_3{ }^2\) then \(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3\) are in
1 \(\mathrm{AP}\)
2 \(\mathrm{HP}\)
3 GP
4 AGP
Explanation:
C : Assume, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be any point on the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) then, \(\mathrm{x}_1^2+\mathrm{y}_1{ }^2=\mathrm{r}_1{ }^2\) Equation of chord of contact of tangents from \(\left(x_1, y_1\right)\) to the circle \(x^2+y^2=r_2^2\) is \(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{r}_2{ }^2\) (ii) touches the circle \(x^2+y^2=r_3{ }^2\) \(\therefore\) The length of perpendicular from centre \((0,0)\) on (ii) is \(=\) radius \(=r_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{r}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \mathrm{r}_2^2=\mathrm{r}_1 \mathrm{r}_3\)Hence, \(\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3\) are in G.P.
120051
If the area of the circle \(7 x^2+7 y^2-7 x+14 y+k\) \(=0\) is \(12 \pi\) sq. units, then the value of \(k\) is
1 \(\frac{-43}{4}\)
2 \(\frac{-301}{4}\)
3 -16
4 \(\pm 4\)
Explanation:
B Equation of circle \(7 x^2+7 y^2-7 x+14 y+k=0\) can be written as, \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+2 \mathrm{y}+\frac{\mathrm{k}}{7}=0\) \(\therefore\) Centre of circle is \(\left(\frac{1}{2},-1\right)\) Radius of circle, \(r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}\) \(=\sqrt{\frac{5}{4}-\frac{\mathrm{k}}{7}} \Rightarrow \mathrm{r}^2=\frac{5}{4}-\frac{\mathrm{k}}{7}\) Given that area of circle \(=12 \pi\) sq. units \(\Rightarrow \pi \mathrm{r}^2=\pi\left(\frac{5}{4}-\frac{\mathrm{k}}{7}\right)=12 \pi\) \(\frac{5}{4}-\frac{\mathrm{k}}{7}=12\) \(\Rightarrow \frac{5}{4}-12=\frac{\mathrm{k}}{7} \Rightarrow \frac{-43}{4}=\frac{\mathrm{k}}{7}\) \(\Rightarrow \mathrm{k}=\frac{-301}{4}\)
COMEDK-2012
Conic Section
120052
If \((-3,2)\) lies on the circle \(x^2+y^2+2 g x+2 f y+c\) \(=0\) which is concentric with the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(6 x+8 y-5=0\), then \(C\) is equal to
1 11
2 -11
3 24
4 100
Explanation:
B If the circle \(x^2+y^2+2 g x+2 f y+c=0\) is concentric with \(x^2+y^2+6 x+8 y-5=0\) then (i) can be written as \(x^2+y^2+6 x+8 y+C=0\) and since it passes through \((-3,2)\), we get \((-3)^2+(2)^2+(6)(-3)+8 \times 2+\mathrm{C}=0\) \(9+4-18+16+\mathrm{C}=0\) \(\mathrm{C}=-11\)
BCECE-2013
Conic Section
120053
The locus of the centers of the circle that are passing through the intersection of the circles \(x^2+y^2=1\) and \(x^2+y^2-2 x+y=0\) is
1 A line whose equation is \(x+2 y=0\)
2 A circle
3 A parabola
4 A line whose equation is \(2 x-y=0\)
Explanation:
A : Let \(S_1=x^2+y^2=1\) \(S_2=x^2+y^2-2 x+y=0\) The equation of a family of circles passing through the intersection of given circle is:- \(S_1+\lambda S_2=0\) \(x^2+y^2-\frac{2 \lambda}{1+\lambda} \mathrm{x}+\frac{\lambda}{1+\lambda} \mathrm{y}-\frac{1}{1+\lambda}=0=0\) Let, \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) be its centre. Then, \(\mathrm{h}=\frac{\lambda}{1+\lambda}\) and \(\mathrm{k}=\) \(-\left(\frac{\lambda}{2(1+\lambda)}\right)\) \(\Rightarrow \mathrm{h}=-2 \mathrm{k}\) Hence, the locus of \((\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}=-2 \mathrm{y}\) \(x+2 y=0\)
APEAPCET-20.08.2021
Conic Section
120054
The point which has the same power with respect to each of the circles \(x^2+y^2-8 x+40=0, x^2+y^2-5 x+16=0\) and \(x^2+y^2-8 x+16 y+160=0\) is
1 \(\left(-8, \frac{-15}{2}\right)\)
2 \(\left(8, \frac{-15}{2}\right)\)
3 \(\left(8, \frac{15}{2}\right)\)
4 \(\left(-8, \frac{15}{2}\right)\)
Explanation:
B Given, \(S_1: x^2+y^2-8 x+40=0\) \(S_2: x^2+y^2-5 x+16=0\) \(S_3: x^2+y^2-8 x+16 y+160=0\) So, \(\mathrm{S}_2-\mathrm{S}_1=0\) \(3 \mathrm{x}-24=0\) \(\mathrm{x}=8\) and, \(S_3-S_1=0\) \(16 y+120=0\) \(4 y+30=0\) \(y=\frac{-15}{2}\) So, Point is \(\left(8, \frac{-15}{2}\right)\) Hence, required point is \(\left(8, \frac{-15}{2}\right)\).
AP EAMCET-19.08.2021
Conic Section
120055
If the chord of contact of tangents from a point on the circle \(x^2+y^2=r_1^2\) to the circle \(x^2+y^2=r_2{ }^2\) touches the circle \(x^2+y^2=r_3{ }^2\) then \(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3\) are in
1 \(\mathrm{AP}\)
2 \(\mathrm{HP}\)
3 GP
4 AGP
Explanation:
C : Assume, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be any point on the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) then, \(\mathrm{x}_1^2+\mathrm{y}_1{ }^2=\mathrm{r}_1{ }^2\) Equation of chord of contact of tangents from \(\left(x_1, y_1\right)\) to the circle \(x^2+y^2=r_2^2\) is \(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{r}_2{ }^2\) (ii) touches the circle \(x^2+y^2=r_3{ }^2\) \(\therefore\) The length of perpendicular from centre \((0,0)\) on (ii) is \(=\) radius \(=r_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{r}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \mathrm{r}_2^2=\mathrm{r}_1 \mathrm{r}_3\)Hence, \(\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3\) are in G.P.
120051
If the area of the circle \(7 x^2+7 y^2-7 x+14 y+k\) \(=0\) is \(12 \pi\) sq. units, then the value of \(k\) is
1 \(\frac{-43}{4}\)
2 \(\frac{-301}{4}\)
3 -16
4 \(\pm 4\)
Explanation:
B Equation of circle \(7 x^2+7 y^2-7 x+14 y+k=0\) can be written as, \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+2 \mathrm{y}+\frac{\mathrm{k}}{7}=0\) \(\therefore\) Centre of circle is \(\left(\frac{1}{2},-1\right)\) Radius of circle, \(r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}\) \(=\sqrt{\frac{5}{4}-\frac{\mathrm{k}}{7}} \Rightarrow \mathrm{r}^2=\frac{5}{4}-\frac{\mathrm{k}}{7}\) Given that area of circle \(=12 \pi\) sq. units \(\Rightarrow \pi \mathrm{r}^2=\pi\left(\frac{5}{4}-\frac{\mathrm{k}}{7}\right)=12 \pi\) \(\frac{5}{4}-\frac{\mathrm{k}}{7}=12\) \(\Rightarrow \frac{5}{4}-12=\frac{\mathrm{k}}{7} \Rightarrow \frac{-43}{4}=\frac{\mathrm{k}}{7}\) \(\Rightarrow \mathrm{k}=\frac{-301}{4}\)
COMEDK-2012
Conic Section
120052
If \((-3,2)\) lies on the circle \(x^2+y^2+2 g x+2 f y+c\) \(=0\) which is concentric with the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(6 x+8 y-5=0\), then \(C\) is equal to
1 11
2 -11
3 24
4 100
Explanation:
B If the circle \(x^2+y^2+2 g x+2 f y+c=0\) is concentric with \(x^2+y^2+6 x+8 y-5=0\) then (i) can be written as \(x^2+y^2+6 x+8 y+C=0\) and since it passes through \((-3,2)\), we get \((-3)^2+(2)^2+(6)(-3)+8 \times 2+\mathrm{C}=0\) \(9+4-18+16+\mathrm{C}=0\) \(\mathrm{C}=-11\)
BCECE-2013
Conic Section
120053
The locus of the centers of the circle that are passing through the intersection of the circles \(x^2+y^2=1\) and \(x^2+y^2-2 x+y=0\) is
1 A line whose equation is \(x+2 y=0\)
2 A circle
3 A parabola
4 A line whose equation is \(2 x-y=0\)
Explanation:
A : Let \(S_1=x^2+y^2=1\) \(S_2=x^2+y^2-2 x+y=0\) The equation of a family of circles passing through the intersection of given circle is:- \(S_1+\lambda S_2=0\) \(x^2+y^2-\frac{2 \lambda}{1+\lambda} \mathrm{x}+\frac{\lambda}{1+\lambda} \mathrm{y}-\frac{1}{1+\lambda}=0=0\) Let, \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) be its centre. Then, \(\mathrm{h}=\frac{\lambda}{1+\lambda}\) and \(\mathrm{k}=\) \(-\left(\frac{\lambda}{2(1+\lambda)}\right)\) \(\Rightarrow \mathrm{h}=-2 \mathrm{k}\) Hence, the locus of \((\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}=-2 \mathrm{y}\) \(x+2 y=0\)
APEAPCET-20.08.2021
Conic Section
120054
The point which has the same power with respect to each of the circles \(x^2+y^2-8 x+40=0, x^2+y^2-5 x+16=0\) and \(x^2+y^2-8 x+16 y+160=0\) is
1 \(\left(-8, \frac{-15}{2}\right)\)
2 \(\left(8, \frac{-15}{2}\right)\)
3 \(\left(8, \frac{15}{2}\right)\)
4 \(\left(-8, \frac{15}{2}\right)\)
Explanation:
B Given, \(S_1: x^2+y^2-8 x+40=0\) \(S_2: x^2+y^2-5 x+16=0\) \(S_3: x^2+y^2-8 x+16 y+160=0\) So, \(\mathrm{S}_2-\mathrm{S}_1=0\) \(3 \mathrm{x}-24=0\) \(\mathrm{x}=8\) and, \(S_3-S_1=0\) \(16 y+120=0\) \(4 y+30=0\) \(y=\frac{-15}{2}\) So, Point is \(\left(8, \frac{-15}{2}\right)\) Hence, required point is \(\left(8, \frac{-15}{2}\right)\).
AP EAMCET-19.08.2021
Conic Section
120055
If the chord of contact of tangents from a point on the circle \(x^2+y^2=r_1^2\) to the circle \(x^2+y^2=r_2{ }^2\) touches the circle \(x^2+y^2=r_3{ }^2\) then \(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3\) are in
1 \(\mathrm{AP}\)
2 \(\mathrm{HP}\)
3 GP
4 AGP
Explanation:
C : Assume, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be any point on the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) then, \(\mathrm{x}_1^2+\mathrm{y}_1{ }^2=\mathrm{r}_1{ }^2\) Equation of chord of contact of tangents from \(\left(x_1, y_1\right)\) to the circle \(x^2+y^2=r_2^2\) is \(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{r}_2{ }^2\) (ii) touches the circle \(x^2+y^2=r_3{ }^2\) \(\therefore\) The length of perpendicular from centre \((0,0)\) on (ii) is \(=\) radius \(=r_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{r}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \mathrm{r}_2^2=\mathrm{r}_1 \mathrm{r}_3\)Hence, \(\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3\) are in G.P.
120051
If the area of the circle \(7 x^2+7 y^2-7 x+14 y+k\) \(=0\) is \(12 \pi\) sq. units, then the value of \(k\) is
1 \(\frac{-43}{4}\)
2 \(\frac{-301}{4}\)
3 -16
4 \(\pm 4\)
Explanation:
B Equation of circle \(7 x^2+7 y^2-7 x+14 y+k=0\) can be written as, \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+2 \mathrm{y}+\frac{\mathrm{k}}{7}=0\) \(\therefore\) Centre of circle is \(\left(\frac{1}{2},-1\right)\) Radius of circle, \(r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}\) \(=\sqrt{\frac{5}{4}-\frac{\mathrm{k}}{7}} \Rightarrow \mathrm{r}^2=\frac{5}{4}-\frac{\mathrm{k}}{7}\) Given that area of circle \(=12 \pi\) sq. units \(\Rightarrow \pi \mathrm{r}^2=\pi\left(\frac{5}{4}-\frac{\mathrm{k}}{7}\right)=12 \pi\) \(\frac{5}{4}-\frac{\mathrm{k}}{7}=12\) \(\Rightarrow \frac{5}{4}-12=\frac{\mathrm{k}}{7} \Rightarrow \frac{-43}{4}=\frac{\mathrm{k}}{7}\) \(\Rightarrow \mathrm{k}=\frac{-301}{4}\)
COMEDK-2012
Conic Section
120052
If \((-3,2)\) lies on the circle \(x^2+y^2+2 g x+2 f y+c\) \(=0\) which is concentric with the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(6 x+8 y-5=0\), then \(C\) is equal to
1 11
2 -11
3 24
4 100
Explanation:
B If the circle \(x^2+y^2+2 g x+2 f y+c=0\) is concentric with \(x^2+y^2+6 x+8 y-5=0\) then (i) can be written as \(x^2+y^2+6 x+8 y+C=0\) and since it passes through \((-3,2)\), we get \((-3)^2+(2)^2+(6)(-3)+8 \times 2+\mathrm{C}=0\) \(9+4-18+16+\mathrm{C}=0\) \(\mathrm{C}=-11\)
BCECE-2013
Conic Section
120053
The locus of the centers of the circle that are passing through the intersection of the circles \(x^2+y^2=1\) and \(x^2+y^2-2 x+y=0\) is
1 A line whose equation is \(x+2 y=0\)
2 A circle
3 A parabola
4 A line whose equation is \(2 x-y=0\)
Explanation:
A : Let \(S_1=x^2+y^2=1\) \(S_2=x^2+y^2-2 x+y=0\) The equation of a family of circles passing through the intersection of given circle is:- \(S_1+\lambda S_2=0\) \(x^2+y^2-\frac{2 \lambda}{1+\lambda} \mathrm{x}+\frac{\lambda}{1+\lambda} \mathrm{y}-\frac{1}{1+\lambda}=0=0\) Let, \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) be its centre. Then, \(\mathrm{h}=\frac{\lambda}{1+\lambda}\) and \(\mathrm{k}=\) \(-\left(\frac{\lambda}{2(1+\lambda)}\right)\) \(\Rightarrow \mathrm{h}=-2 \mathrm{k}\) Hence, the locus of \((\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}=-2 \mathrm{y}\) \(x+2 y=0\)
APEAPCET-20.08.2021
Conic Section
120054
The point which has the same power with respect to each of the circles \(x^2+y^2-8 x+40=0, x^2+y^2-5 x+16=0\) and \(x^2+y^2-8 x+16 y+160=0\) is
1 \(\left(-8, \frac{-15}{2}\right)\)
2 \(\left(8, \frac{-15}{2}\right)\)
3 \(\left(8, \frac{15}{2}\right)\)
4 \(\left(-8, \frac{15}{2}\right)\)
Explanation:
B Given, \(S_1: x^2+y^2-8 x+40=0\) \(S_2: x^2+y^2-5 x+16=0\) \(S_3: x^2+y^2-8 x+16 y+160=0\) So, \(\mathrm{S}_2-\mathrm{S}_1=0\) \(3 \mathrm{x}-24=0\) \(\mathrm{x}=8\) and, \(S_3-S_1=0\) \(16 y+120=0\) \(4 y+30=0\) \(y=\frac{-15}{2}\) So, Point is \(\left(8, \frac{-15}{2}\right)\) Hence, required point is \(\left(8, \frac{-15}{2}\right)\).
AP EAMCET-19.08.2021
Conic Section
120055
If the chord of contact of tangents from a point on the circle \(x^2+y^2=r_1^2\) to the circle \(x^2+y^2=r_2{ }^2\) touches the circle \(x^2+y^2=r_3{ }^2\) then \(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3\) are in
1 \(\mathrm{AP}\)
2 \(\mathrm{HP}\)
3 GP
4 AGP
Explanation:
C : Assume, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be any point on the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) then, \(\mathrm{x}_1^2+\mathrm{y}_1{ }^2=\mathrm{r}_1{ }^2\) Equation of chord of contact of tangents from \(\left(x_1, y_1\right)\) to the circle \(x^2+y^2=r_2^2\) is \(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{r}_2{ }^2\) (ii) touches the circle \(x^2+y^2=r_3{ }^2\) \(\therefore\) The length of perpendicular from centre \((0,0)\) on (ii) is \(=\) radius \(=r_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \frac{\mathrm{r}_2^2}{\sqrt{\mathrm{r}_1^2}}=\mathrm{r}_3\) \(\Rightarrow \mathrm{r}_2^2=\mathrm{r}_1 \mathrm{r}_3\)Hence, \(\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3\) are in G.P.
