TS EAMCET · Maths · Circle
A circle \(C\) passes through \((2 a, 0)\) and the line \(2 x=a\) is the radical axis of the circle \(C\) and the circle \(x^2+y^2=a^2\), then
- A centre of \(C\) is \((-a, 0)\) and \(C\) passes through \((0,0)\) and \(\left(-a_1-a\right)\)
- B circle \(C\) is \(x^2+y^2-2 a x-2 a y=0\)
- C centre of \(C\) is \((a, 0)\) and \(C\) passes through \((0,0)\) and \((a, a)\)
- D centre of \(C\) is \((0,-a)\) and \(C\) passes through \((-a,-a)\) and \((0,0)\)
Answer & Solution
Correct Answer
(C) centre of \(C\) is \((a, 0)\) and \(C\) passes through \((0,0)\) and \((a, a)\)
Step-by-step Solution
Detailed explanation
Radical axis \(P Q\) is \(x-\frac{a}{2}=0\) which cuts the circle \(x^2+y^2=a^2\) in \(P\) and \(Q\). Now, any circle which passes through the points \(P\) and \(Q\) will have radical axis the line \(P Q\) with respect to \(x^2+y^2-a^2=0\). Hence, its equation is the equation of…
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