AP EAMCET · Maths · Circle
From a point P on the circle \(\mathrm{x}^2+\mathrm{y}^2=4\), two tangents are drawn to the circle \(x^2+y^2-6 x-6 y+14=0\). If \(A\) and \(B\) are the points of contact of those lines, then the locus of the centre of the circle passing through the points \(\mathrm{P}, \mathrm{A}\) and B is
- A \(x^2+y^2-3 x-3 y+4=0\)
- B \(2 x^2+2 y^2+6 x+6 y-7=0\)
- C \(x^2+y^2+3 x+3 y-4=0\)
- D \(2 x^2+2 y^2-6 x-6 y+7=0\)
Answer & Solution
Correct Answer
(D) \(2 x^2+2 y^2-6 x-6 y+7=0\)
Step-by-step Solution
Detailed explanation
Center of \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-6 \mathrm{y}+14=0\) is \(O_2=(3,3)\). Point P is \((x_1, y_1)\) on \(\mathrm{x}^2+\mathrm{y}^2=4\), so \(x_1^2+y_1^2=4\). The circle passing through P, A, B has \(PO_2\) as its diameter. Let \((h,k)\) be the center of this…
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