AP EAMCET · Maths · Circle
A circle \(S \equiv x^2+y^2-16=0\) intersects another circle \(S^{\prime}=0\) of radius 5 units such that their common chord is of maximum length. If the slope of that chord is \(\frac{3}{4}\), then the centre of such a circle \(S^{\prime}=0\) is
- A \(\left(\frac{9}{5}, \frac{12}{5}\right)\)
- B \(\left(\frac{5}{9}, \frac{-12}{5}\right)\)
- C \(\left(\frac{-9}{5}, \frac{12}{5}\right)\)
- D \(\left(\frac{3}{5}, \frac{4}{5}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{-9}{5}, \frac{12}{5}\right)\)
Step-by-step Solution
Detailed explanation
\(C_1 = (0,0)\), \(R_1 = 4\) \(R_2 = 5\) Common chord is diameter of smaller circle \(S\), so it passes through \(C_1\). Equation of common chord: \(y - 0 = \frac{3}{4}(x - 0) \implies 3x - 4y = 0\). Line joining centres \(C_1C_2\) is perpendicular to common chord. Slope of…
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