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10-1.Circle and System of Circles
hard
The equation of circle which passes through the point $(1,1)$ and intersect the given circles ${x^2} + {y^2} + 2x + 4y + 6 = 0$ and ${x^2} + {y^2} + 4x + 6y + 2 = 0$ orthogonally, is
A
${x^2} + {y^2} + 16x + 12y + 2 = 0$
B
${x^2} + {y^2} - 16x - 12y - 2 = 0$
C
${x^2} + {y^2} - 16x + 12y + 2 = 0$
D
None of these
Solution
(c) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.
As it intersects orthogonally the given circles,
we have $2g + 4f = 6 + c$ and $4g + 6f = 2 + c$.
As it passes through $(1, 1)$,
we have $2g + 2f = – 2 – c$
From these, we get $g,\;f$ and $c$ as $-8, 6, 2$ respectively and hence equation of circle as
${x^2} + {y^2} – 16x + 12y + 2 = 0$.
Standard 11
Mathematics
