Mathematics - Circle Question with Solution | TestHub

MathematicsCircleFamily of circlesMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

P is a variable point on the line L = 0. Tangents from P touch the circle x² + y² = 4 at Q and R. The parallelogram PQSR is completed.

Tangents from $P$ touch the circle $x^{2} + y^{2} = 4$ at $Q, R$ , and parallelogram $PQSR$ is completed. If $P = (2, 3)$ , then the centre of the circumcircle of $△ QRS$ is

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Answer:

Solution:

$QR$ is $2 x + 3 y = 4$ , and its midpoint is $(8/13, 12/13)$ . Thus $S = Q + R - P = (- 10/13, - 15/13)$ . The circumcircle of $QRS$ is $x^{2} + y^{2} - 4 + λ (2 x + 3 y - 4) = 0$ . Substituting $S$ gives $λ = - 3/13$ ; its centre is (3/13, 9/26).

Stream:JEESubject:MathematicsTopic:CircleSubtopic:Family of circles

⏱ 2mℹ️ Source: QB

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