If the circles x 2 + y 2 + 2 x + 2 k y + 6 = 0 , x 2 + y 2 + 2 k y + k = 0 intersect orthogonally, then k is [2000]

Question

If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ , $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is [2000]

Smart Achievers · Accepted Answer

(1)

Two circles intersect each other orthogonally if $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

Since the two given circles intersect each other orthogonally,

$∴$ $2 (1) (0) + 2 (k) (k) = 6 + k$

$\Rightarrow 2 k^{2} - k - 6 = 0$

$\Rightarrow k = - \frac{3}{2}, 2$

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