TS EAMCET · Maths · Circle
If the circle \(S_1: x^2+y^2=16\) intersects another circle \(S_2\) of radius 5 units such that the common chord is of maximum length and slope \(\frac{3}{4}\), then the centre of the circle \(S_2\) is
- A \(\left(\frac{-9}{5}, \frac{12}{5}\right)\) or \(\left(\frac{9}{5}, \frac{-12}{5}\right)\)
- B \(\left(\frac{7}{5}, \frac{-12}{5}\right)\) or \(\left(\frac{-7}{5}, \frac{12}{5}\right)\)
- C \(\left(\frac{-9}{5}, \frac{-12}{5}\right)\) or \(\left(\frac{9}{5}, \frac{12}{5}\right)\)
- D \(\left(\frac{12}{5}, \frac{9}{5}\right)\) or \(\left(\frac{-12}{5}, \frac{-9}{5}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{-9}{5}, \frac{12}{5}\right)\) or \(\left(\frac{9}{5}, \frac{-12}{5}\right)\)
Step-by-step Solution
Detailed explanation
Given \(S_1=x^2+y^2=16\) Centre \((0,0) r=4\) Clearly, the diameter of \(S_1\) will be common chord. Let \(P Q\) be common chord and centre of \(S_2\) be \((h, k)\) We have \(A P=5, P B=4\), \(\therefore \quad A B=3\) Where \(B=(0,0)\) Slope of \(P Q=\frac{3}{4}\) \(\therefore\)…
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