TS EAMCET · Maths · Circle
A circle C passing through the point \((1,1)\) bisects the circumference of the circle \(x^2+y^2-2 x=0\). If C is orthogonal to the circle \(x^2+y^2+2 y-3=0\) then the centre of the circle C is
- A \(\left(-\frac{1}{2}, 0\right)\)
- B \(\left(\frac{5}{2}, 0\right)\)
- C \(\left(0, \frac{5}{2}\right)\)
- D \(\left(0,-\frac{1}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{5}{2}, 0\right)\)
Step-by-step Solution
Detailed explanation
Let the equation of circle C be \(x^2+y^2+2gx+2fy+c=0\). C passes through \((1,1)\): \(1+1+2g(1)+2f(1)+c=0 \Rightarrow 2g+2f+c+2=0\). Circle \(S_1: x^2+y^2-2x=0\) has center \((1,0)\). C bisects \(S_1\), so the common chord \((x^2+y^2+2gx+2fy+c)-(x^2+y^2-2x)=0\) passes through…
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