TS EAMCET · Maths · Circle
Consider the circle \(x^2+y^2-4 x-2 y+c=0\) whose centre is \(A(2,1)\). If the point \(P(10,7)\) is such that the line segment \(P A\) meets the circle in \(Q\) with \(P Q=5\), then \(c\) is equal to
- A \(-15\)
- B \(20\)
- C \(30\)
- D \(-20\)
Answer & Solution
Correct Answer
(D) \(-20\)
Step-by-step Solution
Detailed explanation
Given equation of circle is \(x^2+y^2-4 x-2 y+c=0\) whose centre is \(A(2,1)\). Now, \(A P=\sqrt{(2-10)^2+(1-7)^2}\) \(\begin{aligned} & =\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=\sqrt{100} \\ & =10\end{aligned}\) \(\therefore \quad A Q=A P-P Q=10-5=5\) So, \(Q\) is the mid-point of AP…
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