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Most Important Selected Qs for JEE AdvancedMathematicsCircle

x^2+y^2=a^2 and (x-2 a)^2+y^2=a^2 are two equal circles touching each other. Find the equation of circle (or circles) of the same radius touching both the circles.

Options

  1. Ax^2+y^2+2 a x+2 3 a y+3 a^2=0
  2. Bx^2+y^2-2 a x+2 3 a y+3 a^2=0
  3. Cx^2+y^2+2 a x-2 3 a y+3 a^2=0
  4. Dx^2+y^2-2 a x-2 3 a y+3 a^2=0

Correct answer

D. x^2+y^2-2 a x-2 3 a y+3 a^2=0

Step-by-step solution

Given circles are x^2+y^2=a^2 (1) and (x-2 a)^2+y^2=a^2 (2) Let A and B be the centres and r₁ and r₂ the radii of the circles (1) and (2) respectively. Then A (0,0), B (2 a, 0), r₁=a, r₂=a Now A B= (0-2 a)^2+0^2 =2 a=r₁+r₂ Hence the two circles touch each other extenally. Let the equation of the circle having same radius 'a' and touching the circles (1) and (2) be (x- )^2+(y- )^2=a^2 (3) Its centre C is ( , ) and radius r₃=a Since circle (3) touches the circle (1), A C=r₁+r₃=2 a [H e r e A C |r₁-r₃ | . as .r₁-r₃=a-

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