MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Tangents from an external point form two congruent right triangles with the radii.The area of the quadrilateral becomes a function of the radius, and we maximise it by equating the factors.Explanation:Given $P(6,8)$ and centre $O(0,0)$.The distance from $P$ to the centre is $OP = \sqrt{6^2 + 8^2} = 10$.Let the radius of the circle be $r$.Since a radius is perpendicular to the tangent at the point of contact, $\triangle OAP$ and $\triangle OBP$ are right triangles.In right triangle $OAP$,$PA = \sqrt{OP^2 - OA^2} = \sqrt{100 - r^2}$.The quadrilateral $PAOB$ consists of two congruent right triangles.So its area is $2 \times \frac{1}{2} \times r \times \sqrt{100 - r^2} = r\sqrt{100 - r^2}$.To maximise this area, maximise its square: $A^2 = r^2(100 - r^2)$.Put $x = r^2$. Then $A^2 = x(100 - x)$.The product $x(100-x)$ is maximum when $x = 100 - x$.Solving gives $2x = 100$, so $x = 50$.Thus $r^2 = 50$, which gives $r = 5\sqrt{2}$.Answer:The radius for which the area of quadrilateral $PAOB$ is maximum is $5\sqrt{2}$, so the correct option is B.