Circles {x^2} + {y^2} + 4x + 6y + 3 = 0 and 2({x^2} + {y^2}) + 6x + 4y + C = 0 will cut orthogonally

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10-1.Circle and System of Circles

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The circles ${x^2} + {y^2} + 4x + 6y + 3 = 0$ and $2({x^2} + {y^2}) + 6x + 4y + C = 0$ will cut orthogonally, if $C$ equals

A

$4$

B

$18$

C

$12$

D

$16$

Solution

(b) $2 \times 2 \times \frac{3}{2} + 2 \times 3 \times 1 = 3 + \frac{C}{2} \Rightarrow C = 18$.

Standard 11

Mathematics

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