If a circle passes through point (1, 2) and cuts circle {x^2} + {y^2} = 4 orthogonally, then equatio

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10-1.Circle and System of Circles

hard

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

$2x + 4y - 1 = 0$

Solution

(c) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$

It passes through $(1, 2)$

$i.e.$ $1 + 4 + 2g + 4f + c = 0$ and is orthogonal w.r.t ${x^2} + {y^2} = 4$.

Therefore, $2g \times 0 + 2f \times 0 = – c + 4$ or $c = 4$

Hence, we get $2g + 4f + 9 = 0$,

then $2x + 4y – 9 = 0$ is the required locus.

Standard 11

Mathematics

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