If the circles x2 + y2 +2x + 2ky + 6 = 0 and x2 + y2 + 2ky + - Maths - Three Dimensional Geometry

Question

If the circles x2 + y2 +2x + 2ky + 6 = 0 and
x2 + y2 + 2ky + k = 0 intersect orthogonally
then k equals ?

Shruti Tyagi · Accepted Answer

Dear Student,

Let R1 and R2 be the radius of both circles and d be the distance between the centres of two circlestwo circles are intersect orthogonally ifR12+R22=d2The equations of circles arex2+y2+2x+2ky+6=0⇒x2+2x+4+y2+2ky+k2+2-k2=0⇒x+22+y+k2=k2-2  --1second equationx2+y2+2ky+k=0x2+y2+2ky+k2-k2+k=0x2+y+k2=k2-k  --2both the equations are converted to genraal form x-h2+y-k2=r2, where h,k is centre of circleHence centres of circles areC1(-2,-k) and C20,-kapply the condition of orthogonal hereR12+R22=d2k2-2+k2-k  =0--22+-k--k22k2-k-2=42k2-k=62k2-k-6=02k2-4k+3k-6=02kk-2+3k-2=0k-22k+3=0hence k=2 and k=-32
Regards,

If the circles x2 + y2 +2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally then k equals ?

If the circles x² + y² +2x + 2ky + 6 = 0 and x² + y² + 2ky + k = 0 intersect orthogonally then k equals ?