If circles {x^2} + {y^2} + 2x + 2ky + 6 = 0 and {x^2} + {y^2} + 2ky + k = 0 intersect orthogonally,

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10-1.Circle and System of Circles

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

A

$2$ or $ - \frac{3}{2}$

B

$-2$ or $\frac{3}{2}$

C

$2$ or $\frac{3}{2}$

D

-$2$ or -$\frac{3}{2}$

(IIT-2000)

Solution

(a) $2gg' + \,2ff' = c + c'\,\,i.e.,\,\,2.1.0 + 2.\,k.\,k. = 6 + k$

or $2{k^2} – k – 6 = 0$ or $(2k + 3)\,(k – 2) = 0$

$\therefore \,$$k = 2,\, – \frac{3}{2}.$

Standard 11

Mathematics

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