AP EAMCET · Maths · Straight Lines
\(A(a, 0)\) is a fixed point and \(\theta\) is a parameter such that \(0 < \theta < 2 \pi\). If \(\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{a} \sin \theta)\) is a point on the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) and \(\mathrm{Q}(\mathrm{b} \sin \theta,-\mathrm{b} \cos \theta)\) is a point on the circle \(x^2+y^2=b^2\), then the locus of the centroid of the triangle APQ is
- A a circle with centre at \(\left(\frac{\mathrm{a}}{3}, 0\right)\) and radius \(\left(\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{3}\right)\)
- B \(a\) circle with centre at \((a, 0)\) and radius \(\left(\frac{\sqrt{a^2+b^2}}{3}\right)\)
- C a parabola with focus at \(\left(\frac{\mathrm{a}}{3}, 0\right)\)
- D a parabola with focus at \((a, 0)\)
Answer & Solution
Correct Answer
(A) a circle with centre at \(\left(\frac{\mathrm{a}}{3}, 0\right)\) and radius \(\left(\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{3}\right)\)
Step-by-step Solution
Detailed explanation
Let \(G(x, y)\) be the centroid of triangle APQ. \(x = \frac{a + a \cos \theta + b \sin \theta}{3}\) \(y = \frac{0 + a \sin \theta - b \cos \theta}{3}\) \(3x - a = a \cos \theta + b \sin \theta\) \(3y = a \sin \theta - b \cos \theta\)…
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