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Orthogonality of Two Circles

The circles ...

Question

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

4

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18

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Solution

The correct option is B $18$
$C_{1} : x^{2} + y^{2} + 4 x + 6 y + 3 = 0$

$g_{1} = 2, f_{1} = 3, c_{1} = 3$

$C_{2} : 2 (x^{2} + y^{2}) + 6 x + 4 y + c = 0$

$g_{2} = \frac{3}{2}, f_{2} = 1, c_{2} = \frac{c}{2}$

As we know that if two circles cut each other orthogonally then they must satisfy-

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

$\Rightarrow 2 \cdot 2 \cdot \frac{3}{2} + 2 \cdot 3 \cdot 1 = 3 + \frac{c}{2}$

$\Rightarrow \frac{c}{2} = 6 + 6 - 3$

$\Rightarrow c = 9 \times 2 = 18$

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