If a circle passes through the point $(1, 2)$ and cuts the circle ${x^2} + {y^2} = 4$ orthogonally, then the equation of the locus of its centre is
If a circle passes through the point $(1, 2)$ and cuts the circle ${x^2} + {y^2} = 4$ orthogonally, then the equation of the locus of its centre is
- A
${x^2} + {y^2} - 3x - 8y + 1 = 0$
- B
${x^2} + {y^2} - 2x - 6y - 7 = 0$
- C
$2x + 4y - 9 = 0$
- D
$2x + 4y - 1 = 0$
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The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is
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- [JEE MAIN 2015]
The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be
The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be
Number of common tangents to the circles
$x^2 + y^2 -2x + 4y -4 = 0$ and
$x^2 + y^2 -8x -4y + 16 = 0 $ is-
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The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that
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