Prove that the circles x2+y2+24ux+2vy=0 and x2+y2+2u1x+2v1y=0 touch each other if 12uv1 = u1v - Maths - Conic Sections

Question

Prove that the circles x 2 + y 2 + 24 u x + 2 v y = 0   and
  x 2 + y 2 + 2 u 1 x + 2 v 1 y = 0 touch each other if
12uv 1 = u 1 v

Neha Sethi · Accepted Answer

Dear student
To show circles x2+y2+24ux+2vy=0 and x2+y2+2u1x+2v1y=0 touch other
Given: 12uv1=u1v⇒12u=u1vv1   1i e
 To show: C1C2=r1+r2 , where C1 and C2 are the centres of the  circles  respectivelyand r1 and r2 are the radius of the circles respectively
Consider, x2+y2+24ux+2vy=0Here, g1=12u and f1=v and c1=0radius, r1=g12+f12-c1=12u2+v2-0=144u2+v2Centre , C1=-g1,-f1=-12u,-vSimilarly consider,  x2+y2+2u1x+2v1y=0 Here, g2=u1 and f2=v1 and c2=0radius, r2=g22+f22-c2=u12+v12-0=u12+v12Centre , C2=-g2,-f2=-u1,-v1 Now, C1C2=-u1+12u2+(-v1+v)2   using distance formulaC1C2=u12+144u2-24uu1+v12+v22-2vv1C1C2=144u2+v2+u12+v12-212uu1+vv1C1C2=r12+r22-212uu1+vv1C1C2=r12+r22-2u12vv1+vv1C1C2=r12+r22-2u12v+vv12v1C1C2=r12+r22-2vu12+v12v1C1C2=r12+r22-2vr22v1
Please recheck the question there seems to be some mistake in it as
it cannot be solved further to get the desired result
Regards

Prove that the circles x 2 + y 2 + 24 u x + 2 v y = 0 and x 2 + y 2 + 2 u 1 x + 2 v 1 y = 0 touch each other if 12uv 1 = u 1 v.

Prove that the circles $x^{2} + y^{2} + 24 u x + 2 v y = 0 and x^{2} + y^{2} + 2 u_{1} x + 2 v_{1} y = 0$ touch each other if 12uv ₁ = u ₁ v.