120004
If \(x^2+y^2+6 x+2 k y+25=0\) to touch \(Y\)-axis, then \(\mathrm{k}=\)
1 \(\pm 20\)
2 \(-1,-5\)
3 \(\pm 5\)
4 4
Explanation:
C Given circle \(\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+2 \mathrm{ky}+25=0\) touches \(\mathrm{Y}\)-axis and here centre is \((-3,-\mathrm{k})\) so radius \(=3\) units Also radius \(=\sqrt{9+\mathrm{k}^2-25}\) \(\Rightarrow \sqrt{\mathrm{k}^2-16}=3 \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
AP EAMCET-17.09.2020
Conic Section
120005
If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\) then the other end of the diameter is
1 \((0,1)\)
2 \((1,1)\)
3 \((1,2)\)
4 \((1,0)\)
Explanation:
C Given, circle \(x^2+y^2-4 x-6 y+11=0\) Center \(=(2,3)\) One end of diameter \(=(3,4)\) Let, other end be \((\mathrm{h}, \mathrm{k})\) So, \(\quad \frac{\mathrm{h}+3}{2}=2, \frac{\mathrm{k}+4}{2}=3\) \(\mathrm{h}=1, \mathrm{k}=2\) Hence, \((\mathrm{h}, \mathrm{k})=(1,2)\)
AP EAMCET-17.09.2020
Conic Section
120014
The common tangent to the circles \(x^2+y^2=4\) and \(x^2+y^2+6 x+8 y-24=0\) also passes through the point
1 \((6,-2)\)
2 \((4,-2)\)
3 \((-6,4)\)
4 \((-4,6)\)
Explanation:
A Given two circle equation, \(S_1=x^2+y^2=4\) With centre \(C_1=(0,0), r_1=2\) \(\mathrm{S}_2 =\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+8 \mathrm{y}-24=0\) \((\mathrm{x} +3)^2+(\mathrm{y}+4)^2=49\) \(\mathrm{C}_2 =(-3,-4), \mathrm{r}_2=7\) \(\mathrm{C}_1 \mathrm{C}_2 =\left|\mathrm{r}_1-\mathrm{r}_2\right|\) Equation of common tangent \(\mathrm{S}_1-\mathrm{S}_2=0\) \(6 x+8 y-20=0\) \(3 x+4 y=10\)Hence, \((6,-2)\) satisfies the equation.
JEE Main 09.04.2019
Conic Section
119978
The locus of the centre of a circle which touches externally the circle \(x^2+y^2-6 x-6 y+\) \(14=0\) and also touches the \(y\)-axis is given by the equation
1 \(x^2-6 x-10 y+14=0\)
2 \(x^2-10 x-6 y+14=0\)
3 \(y^2-6 x-10 y+14=0\)
4 \(y^2-10 x-6 y+14=0\)
Explanation:
D Given, \(x^2+y^2-6 x-6 y+14=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { by comparison of both equations }\) \(2 \mathrm{~g}=-6, \quad 2 \mathrm{f}=-6, \mathrm{C}=14\) \(\mathrm{g}=-3,\) \((-\mathrm{g},-\mathrm{f})=(3,3), \mathrm{f}=-3,\) \(\mathrm{C}=14\) \(r=\sqrt{(-3)^2+(-3)^2-14}\) \(=\sqrt{9+9-14}=\sqrt{4}=2\) \(\therefore \mathrm{C}_1 \mathrm{C}_2=\mathrm{h}+2\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(\mathrm{h}-3)^2+(\mathrm{k}-3)^2}\) \(\mathrm{~h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\) We know that, From equation (i) and equation (ii), we get - \(\mathrm{h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\)
JEEE-2018
Conic Section
119979
If the circles \(x^2+y^2+2 x+2 k y+6=0, x^2+y^2+\) \(2 \mathrm{ky}+\mathrm{k}=\mathbf{0}\) intersect orthogonally, then \(\mathrm{k}\) : is
1 2 (or) \(\frac{-3}{2}\)
2 -2 (or) \(\frac{-3}{2}\)
3 2 (or) \(\frac{3}{2}\)
4 -2 (or) \(\frac{3}{2}\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{ky}+6=0\) \(\ldots .\). (i)\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{ky}+\mathrm{k}=0\) \(\ldots .\). (ii)\) We know that, General equation of a circle, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) For equation (i), we get - \(2 \mathrm{~g}_1=2, \quad\), \(\mathrm{f}=2 \mathrm{k}\), \(\mathrm{c}_1=6\) \(\mathrm{~g}_1=1, \mathrm{f}_1=\mathrm{k} \quad\) For equation (ii), we get - \(2 \mathrm{~g}_2=0, \quad 2 \mathrm{f}_2=2 \mathrm{k}\), \(\mathrm{c}_2=\mathrm{k}\) \(\mathrm{g}_2=0 \quad \quad \mathrm{f}_2=\mathrm{k}\) \(\mathrm{c}_2=\mathrm{k}\) Using condition of orthoganality \(2 \mathrm{f}_1 \mathrm{f}_2+2 \mathrm{~g}_1 \mathrm{~g}_2=\mathrm{c}_1+\mathrm{c}_2\) \(2 \times \mathrm{k} \times \mathrm{k}+2 \times 1 \times 0=6+\mathrm{k}\) \(2 \mathrm{k}^2=6+\mathrm{k}\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-4 \mathrm{k}+3 \mathrm{k}-6=0\) \(2 \mathrm{k}(\mathrm{k}-2)+3(\mathrm{k}-2)=0\) \(\mathrm{k}=2, \mathrm{k}=\frac{-3}{2}\) \(\mathrm{k}=2 \text { or } \frac{-3}{2}\)
120004
If \(x^2+y^2+6 x+2 k y+25=0\) to touch \(Y\)-axis, then \(\mathrm{k}=\)
1 \(\pm 20\)
2 \(-1,-5\)
3 \(\pm 5\)
4 4
Explanation:
C Given circle \(\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+2 \mathrm{ky}+25=0\) touches \(\mathrm{Y}\)-axis and here centre is \((-3,-\mathrm{k})\) so radius \(=3\) units Also radius \(=\sqrt{9+\mathrm{k}^2-25}\) \(\Rightarrow \sqrt{\mathrm{k}^2-16}=3 \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
AP EAMCET-17.09.2020
Conic Section
120005
If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\) then the other end of the diameter is
1 \((0,1)\)
2 \((1,1)\)
3 \((1,2)\)
4 \((1,0)\)
Explanation:
C Given, circle \(x^2+y^2-4 x-6 y+11=0\) Center \(=(2,3)\) One end of diameter \(=(3,4)\) Let, other end be \((\mathrm{h}, \mathrm{k})\) So, \(\quad \frac{\mathrm{h}+3}{2}=2, \frac{\mathrm{k}+4}{2}=3\) \(\mathrm{h}=1, \mathrm{k}=2\) Hence, \((\mathrm{h}, \mathrm{k})=(1,2)\)
AP EAMCET-17.09.2020
Conic Section
120014
The common tangent to the circles \(x^2+y^2=4\) and \(x^2+y^2+6 x+8 y-24=0\) also passes through the point
1 \((6,-2)\)
2 \((4,-2)\)
3 \((-6,4)\)
4 \((-4,6)\)
Explanation:
A Given two circle equation, \(S_1=x^2+y^2=4\) With centre \(C_1=(0,0), r_1=2\) \(\mathrm{S}_2 =\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+8 \mathrm{y}-24=0\) \((\mathrm{x} +3)^2+(\mathrm{y}+4)^2=49\) \(\mathrm{C}_2 =(-3,-4), \mathrm{r}_2=7\) \(\mathrm{C}_1 \mathrm{C}_2 =\left|\mathrm{r}_1-\mathrm{r}_2\right|\) Equation of common tangent \(\mathrm{S}_1-\mathrm{S}_2=0\) \(6 x+8 y-20=0\) \(3 x+4 y=10\)Hence, \((6,-2)\) satisfies the equation.
JEE Main 09.04.2019
Conic Section
119978
The locus of the centre of a circle which touches externally the circle \(x^2+y^2-6 x-6 y+\) \(14=0\) and also touches the \(y\)-axis is given by the equation
1 \(x^2-6 x-10 y+14=0\)
2 \(x^2-10 x-6 y+14=0\)
3 \(y^2-6 x-10 y+14=0\)
4 \(y^2-10 x-6 y+14=0\)
Explanation:
D Given, \(x^2+y^2-6 x-6 y+14=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { by comparison of both equations }\) \(2 \mathrm{~g}=-6, \quad 2 \mathrm{f}=-6, \mathrm{C}=14\) \(\mathrm{g}=-3,\) \((-\mathrm{g},-\mathrm{f})=(3,3), \mathrm{f}=-3,\) \(\mathrm{C}=14\) \(r=\sqrt{(-3)^2+(-3)^2-14}\) \(=\sqrt{9+9-14}=\sqrt{4}=2\) \(\therefore \mathrm{C}_1 \mathrm{C}_2=\mathrm{h}+2\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(\mathrm{h}-3)^2+(\mathrm{k}-3)^2}\) \(\mathrm{~h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\) We know that, From equation (i) and equation (ii), we get - \(\mathrm{h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\)
JEEE-2018
Conic Section
119979
If the circles \(x^2+y^2+2 x+2 k y+6=0, x^2+y^2+\) \(2 \mathrm{ky}+\mathrm{k}=\mathbf{0}\) intersect orthogonally, then \(\mathrm{k}\) : is
1 2 (or) \(\frac{-3}{2}\)
2 -2 (or) \(\frac{-3}{2}\)
3 2 (or) \(\frac{3}{2}\)
4 -2 (or) \(\frac{3}{2}\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{ky}+6=0\) \(\ldots .\). (i)\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{ky}+\mathrm{k}=0\) \(\ldots .\). (ii)\) We know that, General equation of a circle, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) For equation (i), we get - \(2 \mathrm{~g}_1=2, \quad\), \(\mathrm{f}=2 \mathrm{k}\), \(\mathrm{c}_1=6\) \(\mathrm{~g}_1=1, \mathrm{f}_1=\mathrm{k} \quad\) For equation (ii), we get - \(2 \mathrm{~g}_2=0, \quad 2 \mathrm{f}_2=2 \mathrm{k}\), \(\mathrm{c}_2=\mathrm{k}\) \(\mathrm{g}_2=0 \quad \quad \mathrm{f}_2=\mathrm{k}\) \(\mathrm{c}_2=\mathrm{k}\) Using condition of orthoganality \(2 \mathrm{f}_1 \mathrm{f}_2+2 \mathrm{~g}_1 \mathrm{~g}_2=\mathrm{c}_1+\mathrm{c}_2\) \(2 \times \mathrm{k} \times \mathrm{k}+2 \times 1 \times 0=6+\mathrm{k}\) \(2 \mathrm{k}^2=6+\mathrm{k}\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-4 \mathrm{k}+3 \mathrm{k}-6=0\) \(2 \mathrm{k}(\mathrm{k}-2)+3(\mathrm{k}-2)=0\) \(\mathrm{k}=2, \mathrm{k}=\frac{-3}{2}\) \(\mathrm{k}=2 \text { or } \frac{-3}{2}\)
120004
If \(x^2+y^2+6 x+2 k y+25=0\) to touch \(Y\)-axis, then \(\mathrm{k}=\)
1 \(\pm 20\)
2 \(-1,-5\)
3 \(\pm 5\)
4 4
Explanation:
C Given circle \(\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+2 \mathrm{ky}+25=0\) touches \(\mathrm{Y}\)-axis and here centre is \((-3,-\mathrm{k})\) so radius \(=3\) units Also radius \(=\sqrt{9+\mathrm{k}^2-25}\) \(\Rightarrow \sqrt{\mathrm{k}^2-16}=3 \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
AP EAMCET-17.09.2020
Conic Section
120005
If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\) then the other end of the diameter is
1 \((0,1)\)
2 \((1,1)\)
3 \((1,2)\)
4 \((1,0)\)
Explanation:
C Given, circle \(x^2+y^2-4 x-6 y+11=0\) Center \(=(2,3)\) One end of diameter \(=(3,4)\) Let, other end be \((\mathrm{h}, \mathrm{k})\) So, \(\quad \frac{\mathrm{h}+3}{2}=2, \frac{\mathrm{k}+4}{2}=3\) \(\mathrm{h}=1, \mathrm{k}=2\) Hence, \((\mathrm{h}, \mathrm{k})=(1,2)\)
AP EAMCET-17.09.2020
Conic Section
120014
The common tangent to the circles \(x^2+y^2=4\) and \(x^2+y^2+6 x+8 y-24=0\) also passes through the point
1 \((6,-2)\)
2 \((4,-2)\)
3 \((-6,4)\)
4 \((-4,6)\)
Explanation:
A Given two circle equation, \(S_1=x^2+y^2=4\) With centre \(C_1=(0,0), r_1=2\) \(\mathrm{S}_2 =\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+8 \mathrm{y}-24=0\) \((\mathrm{x} +3)^2+(\mathrm{y}+4)^2=49\) \(\mathrm{C}_2 =(-3,-4), \mathrm{r}_2=7\) \(\mathrm{C}_1 \mathrm{C}_2 =\left|\mathrm{r}_1-\mathrm{r}_2\right|\) Equation of common tangent \(\mathrm{S}_1-\mathrm{S}_2=0\) \(6 x+8 y-20=0\) \(3 x+4 y=10\)Hence, \((6,-2)\) satisfies the equation.
JEE Main 09.04.2019
Conic Section
119978
The locus of the centre of a circle which touches externally the circle \(x^2+y^2-6 x-6 y+\) \(14=0\) and also touches the \(y\)-axis is given by the equation
1 \(x^2-6 x-10 y+14=0\)
2 \(x^2-10 x-6 y+14=0\)
3 \(y^2-6 x-10 y+14=0\)
4 \(y^2-10 x-6 y+14=0\)
Explanation:
D Given, \(x^2+y^2-6 x-6 y+14=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { by comparison of both equations }\) \(2 \mathrm{~g}=-6, \quad 2 \mathrm{f}=-6, \mathrm{C}=14\) \(\mathrm{g}=-3,\) \((-\mathrm{g},-\mathrm{f})=(3,3), \mathrm{f}=-3,\) \(\mathrm{C}=14\) \(r=\sqrt{(-3)^2+(-3)^2-14}\) \(=\sqrt{9+9-14}=\sqrt{4}=2\) \(\therefore \mathrm{C}_1 \mathrm{C}_2=\mathrm{h}+2\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(\mathrm{h}-3)^2+(\mathrm{k}-3)^2}\) \(\mathrm{~h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\) We know that, From equation (i) and equation (ii), we get - \(\mathrm{h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\)
JEEE-2018
Conic Section
119979
If the circles \(x^2+y^2+2 x+2 k y+6=0, x^2+y^2+\) \(2 \mathrm{ky}+\mathrm{k}=\mathbf{0}\) intersect orthogonally, then \(\mathrm{k}\) : is
1 2 (or) \(\frac{-3}{2}\)
2 -2 (or) \(\frac{-3}{2}\)
3 2 (or) \(\frac{3}{2}\)
4 -2 (or) \(\frac{3}{2}\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{ky}+6=0\) \(\ldots .\). (i)\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{ky}+\mathrm{k}=0\) \(\ldots .\). (ii)\) We know that, General equation of a circle, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) For equation (i), we get - \(2 \mathrm{~g}_1=2, \quad\), \(\mathrm{f}=2 \mathrm{k}\), \(\mathrm{c}_1=6\) \(\mathrm{~g}_1=1, \mathrm{f}_1=\mathrm{k} \quad\) For equation (ii), we get - \(2 \mathrm{~g}_2=0, \quad 2 \mathrm{f}_2=2 \mathrm{k}\), \(\mathrm{c}_2=\mathrm{k}\) \(\mathrm{g}_2=0 \quad \quad \mathrm{f}_2=\mathrm{k}\) \(\mathrm{c}_2=\mathrm{k}\) Using condition of orthoganality \(2 \mathrm{f}_1 \mathrm{f}_2+2 \mathrm{~g}_1 \mathrm{~g}_2=\mathrm{c}_1+\mathrm{c}_2\) \(2 \times \mathrm{k} \times \mathrm{k}+2 \times 1 \times 0=6+\mathrm{k}\) \(2 \mathrm{k}^2=6+\mathrm{k}\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-4 \mathrm{k}+3 \mathrm{k}-6=0\) \(2 \mathrm{k}(\mathrm{k}-2)+3(\mathrm{k}-2)=0\) \(\mathrm{k}=2, \mathrm{k}=\frac{-3}{2}\) \(\mathrm{k}=2 \text { or } \frac{-3}{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
120004
If \(x^2+y^2+6 x+2 k y+25=0\) to touch \(Y\)-axis, then \(\mathrm{k}=\)
1 \(\pm 20\)
2 \(-1,-5\)
3 \(\pm 5\)
4 4
Explanation:
C Given circle \(\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+2 \mathrm{ky}+25=0\) touches \(\mathrm{Y}\)-axis and here centre is \((-3,-\mathrm{k})\) so radius \(=3\) units Also radius \(=\sqrt{9+\mathrm{k}^2-25}\) \(\Rightarrow \sqrt{\mathrm{k}^2-16}=3 \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
AP EAMCET-17.09.2020
Conic Section
120005
If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\) then the other end of the diameter is
1 \((0,1)\)
2 \((1,1)\)
3 \((1,2)\)
4 \((1,0)\)
Explanation:
C Given, circle \(x^2+y^2-4 x-6 y+11=0\) Center \(=(2,3)\) One end of diameter \(=(3,4)\) Let, other end be \((\mathrm{h}, \mathrm{k})\) So, \(\quad \frac{\mathrm{h}+3}{2}=2, \frac{\mathrm{k}+4}{2}=3\) \(\mathrm{h}=1, \mathrm{k}=2\) Hence, \((\mathrm{h}, \mathrm{k})=(1,2)\)
AP EAMCET-17.09.2020
Conic Section
120014
The common tangent to the circles \(x^2+y^2=4\) and \(x^2+y^2+6 x+8 y-24=0\) also passes through the point
1 \((6,-2)\)
2 \((4,-2)\)
3 \((-6,4)\)
4 \((-4,6)\)
Explanation:
A Given two circle equation, \(S_1=x^2+y^2=4\) With centre \(C_1=(0,0), r_1=2\) \(\mathrm{S}_2 =\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+8 \mathrm{y}-24=0\) \((\mathrm{x} +3)^2+(\mathrm{y}+4)^2=49\) \(\mathrm{C}_2 =(-3,-4), \mathrm{r}_2=7\) \(\mathrm{C}_1 \mathrm{C}_2 =\left|\mathrm{r}_1-\mathrm{r}_2\right|\) Equation of common tangent \(\mathrm{S}_1-\mathrm{S}_2=0\) \(6 x+8 y-20=0\) \(3 x+4 y=10\)Hence, \((6,-2)\) satisfies the equation.
JEE Main 09.04.2019
Conic Section
119978
The locus of the centre of a circle which touches externally the circle \(x^2+y^2-6 x-6 y+\) \(14=0\) and also touches the \(y\)-axis is given by the equation
1 \(x^2-6 x-10 y+14=0\)
2 \(x^2-10 x-6 y+14=0\)
3 \(y^2-6 x-10 y+14=0\)
4 \(y^2-10 x-6 y+14=0\)
Explanation:
D Given, \(x^2+y^2-6 x-6 y+14=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { by comparison of both equations }\) \(2 \mathrm{~g}=-6, \quad 2 \mathrm{f}=-6, \mathrm{C}=14\) \(\mathrm{g}=-3,\) \((-\mathrm{g},-\mathrm{f})=(3,3), \mathrm{f}=-3,\) \(\mathrm{C}=14\) \(r=\sqrt{(-3)^2+(-3)^2-14}\) \(=\sqrt{9+9-14}=\sqrt{4}=2\) \(\therefore \mathrm{C}_1 \mathrm{C}_2=\mathrm{h}+2\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(\mathrm{h}-3)^2+(\mathrm{k}-3)^2}\) \(\mathrm{~h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\) We know that, From equation (i) and equation (ii), we get - \(\mathrm{h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\)
JEEE-2018
Conic Section
119979
If the circles \(x^2+y^2+2 x+2 k y+6=0, x^2+y^2+\) \(2 \mathrm{ky}+\mathrm{k}=\mathbf{0}\) intersect orthogonally, then \(\mathrm{k}\) : is
1 2 (or) \(\frac{-3}{2}\)
2 -2 (or) \(\frac{-3}{2}\)
3 2 (or) \(\frac{3}{2}\)
4 -2 (or) \(\frac{3}{2}\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{ky}+6=0\) \(\ldots .\). (i)\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{ky}+\mathrm{k}=0\) \(\ldots .\). (ii)\) We know that, General equation of a circle, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) For equation (i), we get - \(2 \mathrm{~g}_1=2, \quad\), \(\mathrm{f}=2 \mathrm{k}\), \(\mathrm{c}_1=6\) \(\mathrm{~g}_1=1, \mathrm{f}_1=\mathrm{k} \quad\) For equation (ii), we get - \(2 \mathrm{~g}_2=0, \quad 2 \mathrm{f}_2=2 \mathrm{k}\), \(\mathrm{c}_2=\mathrm{k}\) \(\mathrm{g}_2=0 \quad \quad \mathrm{f}_2=\mathrm{k}\) \(\mathrm{c}_2=\mathrm{k}\) Using condition of orthoganality \(2 \mathrm{f}_1 \mathrm{f}_2+2 \mathrm{~g}_1 \mathrm{~g}_2=\mathrm{c}_1+\mathrm{c}_2\) \(2 \times \mathrm{k} \times \mathrm{k}+2 \times 1 \times 0=6+\mathrm{k}\) \(2 \mathrm{k}^2=6+\mathrm{k}\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-4 \mathrm{k}+3 \mathrm{k}-6=0\) \(2 \mathrm{k}(\mathrm{k}-2)+3(\mathrm{k}-2)=0\) \(\mathrm{k}=2, \mathrm{k}=\frac{-3}{2}\) \(\mathrm{k}=2 \text { or } \frac{-3}{2}\)
120004
If \(x^2+y^2+6 x+2 k y+25=0\) to touch \(Y\)-axis, then \(\mathrm{k}=\)
1 \(\pm 20\)
2 \(-1,-5\)
3 \(\pm 5\)
4 4
Explanation:
C Given circle \(\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+2 \mathrm{ky}+25=0\) touches \(\mathrm{Y}\)-axis and here centre is \((-3,-\mathrm{k})\) so radius \(=3\) units Also radius \(=\sqrt{9+\mathrm{k}^2-25}\) \(\Rightarrow \sqrt{\mathrm{k}^2-16}=3 \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
AP EAMCET-17.09.2020
Conic Section
120005
If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\) then the other end of the diameter is
1 \((0,1)\)
2 \((1,1)\)
3 \((1,2)\)
4 \((1,0)\)
Explanation:
C Given, circle \(x^2+y^2-4 x-6 y+11=0\) Center \(=(2,3)\) One end of diameter \(=(3,4)\) Let, other end be \((\mathrm{h}, \mathrm{k})\) So, \(\quad \frac{\mathrm{h}+3}{2}=2, \frac{\mathrm{k}+4}{2}=3\) \(\mathrm{h}=1, \mathrm{k}=2\) Hence, \((\mathrm{h}, \mathrm{k})=(1,2)\)
AP EAMCET-17.09.2020
Conic Section
120014
The common tangent to the circles \(x^2+y^2=4\) and \(x^2+y^2+6 x+8 y-24=0\) also passes through the point
1 \((6,-2)\)
2 \((4,-2)\)
3 \((-6,4)\)
4 \((-4,6)\)
Explanation:
A Given two circle equation, \(S_1=x^2+y^2=4\) With centre \(C_1=(0,0), r_1=2\) \(\mathrm{S}_2 =\mathrm{x}^2+\mathrm{y}^2+6 \mathrm{x}+8 \mathrm{y}-24=0\) \((\mathrm{x} +3)^2+(\mathrm{y}+4)^2=49\) \(\mathrm{C}_2 =(-3,-4), \mathrm{r}_2=7\) \(\mathrm{C}_1 \mathrm{C}_2 =\left|\mathrm{r}_1-\mathrm{r}_2\right|\) Equation of common tangent \(\mathrm{S}_1-\mathrm{S}_2=0\) \(6 x+8 y-20=0\) \(3 x+4 y=10\)Hence, \((6,-2)\) satisfies the equation.
JEE Main 09.04.2019
Conic Section
119978
The locus of the centre of a circle which touches externally the circle \(x^2+y^2-6 x-6 y+\) \(14=0\) and also touches the \(y\)-axis is given by the equation
1 \(x^2-6 x-10 y+14=0\)
2 \(x^2-10 x-6 y+14=0\)
3 \(y^2-6 x-10 y+14=0\)
4 \(y^2-10 x-6 y+14=0\)
Explanation:
D Given, \(x^2+y^2-6 x-6 y+14=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { by comparison of both equations }\) \(2 \mathrm{~g}=-6, \quad 2 \mathrm{f}=-6, \mathrm{C}=14\) \(\mathrm{g}=-3,\) \((-\mathrm{g},-\mathrm{f})=(3,3), \mathrm{f}=-3,\) \(\mathrm{C}=14\) \(r=\sqrt{(-3)^2+(-3)^2-14}\) \(=\sqrt{9+9-14}=\sqrt{4}=2\) \(\therefore \mathrm{C}_1 \mathrm{C}_2=\mathrm{h}+2\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(\mathrm{h}-3)^2+(\mathrm{k}-3)^2}\) \(\mathrm{~h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\) We know that, From equation (i) and equation (ii), we get - \(\mathrm{h}+2=\sqrt{(\mathrm{h}-3)^3+(\mathrm{k}-3)^2}\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=(\mathrm{h}-3)^2+(\mathrm{k}-3)^2\) \(\mathrm{~h}^2+4+4 \mathrm{~h}=\mathrm{h}^2+9-6 \mathrm{~h}+\mathrm{k}^2+9-6 \mathrm{k}\) \(\mathrm{k}^2-10 \mathrm{~h}-6 \mathrm{k}+14=0\) \(\mathrm{y}^2-10 \mathrm{x}-6 \mathrm{y}+14=0\)
JEEE-2018
Conic Section
119979
If the circles \(x^2+y^2+2 x+2 k y+6=0, x^2+y^2+\) \(2 \mathrm{ky}+\mathrm{k}=\mathbf{0}\) intersect orthogonally, then \(\mathrm{k}\) : is
1 2 (or) \(\frac{-3}{2}\)
2 -2 (or) \(\frac{-3}{2}\)
3 2 (or) \(\frac{3}{2}\)
4 -2 (or) \(\frac{3}{2}\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{ky}+6=0\) \(\ldots .\). (i)\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{ky}+\mathrm{k}=0\) \(\ldots .\). (ii)\) We know that, General equation of a circle, \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) For equation (i), we get - \(2 \mathrm{~g}_1=2, \quad\), \(\mathrm{f}=2 \mathrm{k}\), \(\mathrm{c}_1=6\) \(\mathrm{~g}_1=1, \mathrm{f}_1=\mathrm{k} \quad\) For equation (ii), we get - \(2 \mathrm{~g}_2=0, \quad 2 \mathrm{f}_2=2 \mathrm{k}\), \(\mathrm{c}_2=\mathrm{k}\) \(\mathrm{g}_2=0 \quad \quad \mathrm{f}_2=\mathrm{k}\) \(\mathrm{c}_2=\mathrm{k}\) Using condition of orthoganality \(2 \mathrm{f}_1 \mathrm{f}_2+2 \mathrm{~g}_1 \mathrm{~g}_2=\mathrm{c}_1+\mathrm{c}_2\) \(2 \times \mathrm{k} \times \mathrm{k}+2 \times 1 \times 0=6+\mathrm{k}\) \(2 \mathrm{k}^2=6+\mathrm{k}\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-\mathrm{k}-6=0\) \(2 \mathrm{k}^2-4 \mathrm{k}+3 \mathrm{k}-6=0\) \(2 \mathrm{k}(\mathrm{k}-2)+3(\mathrm{k}-2)=0\) \(\mathrm{k}=2, \mathrm{k}=\frac{-3}{2}\) \(\mathrm{k}=2 \text { or } \frac{-3}{2}\)
