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AP EAMCET · Maths · Circle

The locus of centres of all circles which touch the line \(x=2 a\) and cut the circle \(x^2+y^2=a^2\) orthogonally is

  1. A \(y^2+4 a x-5 a^2=0\)
  2. B \(y^2+4 a x+5 a^2=0\)
  3. C \(y^2=4 a x-5 a^2\)
  4. D \(y^2=4 a x+5 a^2\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(y^2+4 a x-5 a^2=0\)

Step-by-step Solution

Detailed explanation

Let the center of the circle be \((h, k)\) and its radius be \(r\). Condition for touching the line \(x=2a\): \(r = |h-2a|\) \(r^2 = (h-2a)^2\) Condition for orthogonal intersection with \(x^2+y^2=a^2\): \(2(-h)(0) + 2(-k)(0) = (h^2+k^2-r^2) + (-a^2)\) \(0 = h^2+k^2-r^2-a^2\)…
From AP EAMCET
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