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TS EAMCET · Maths · Circle

The centre of the circle passing through the points of intersection of the circles \((x+3)^2+(y+2)^2=25\) and \((x-2)^2+(y-3)^2=25\) and cutting the circle \((x+1)^2+(y-2)^2=16\) orthogonally is

  1. A \(\left(\frac{-27}{2}, \frac{-25}{2}\right)\)
  2. B \((0,0)\)
  3. C \(\left(\frac{16}{3}, \frac{-25}{4}\right)\)
  4. D \(\left(\frac{4}{7}, \frac{3}{7}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(\frac{-27}{2}, \frac{-25}{2}\right)\)

Step-by-step Solution

Detailed explanation

Equation of circle passes through the point of intersections of given circles \(S_1 \equiv(x+3)^2+(y+2)^2=25\) and \(\quad S_2=(x-2)^2+(y-3)^2=25\) is \(S_1+\lambda S_2=0\) \(\Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+(6-4 \lambda) x+(4-6 \lambda) y\) \(-12(1+\lambda)=0\)…
From TS EAMCET
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