JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q73

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MathematicsSingle correct+4 / -1Question 73 of 90

If a point P has co-ordinates (0, −2) and Q is any point on the circle, $x^2+y^2-5x-y+5=0$ , then the maximum value of $(PQ)^2$ is :

\dfrac{25+\sqrt6}2

14+5\sqrt3

\dfrac{47+10\sqrt6}2

8+5\sqrt3

Answer and worked solution

Official answer: B

The circle has centre $\left(\frac52,\frac12\right)$ and $r=\frac{\sqrt6}2$ . The distance from P is $\frac5{\sqrt2}$ , so
$PQ_{max}^2=\left(\frac5{\sqrt2}+\frac{\sqrt6}2\right)^2=\frac{25}2+5\sqrt3+\frac32=14+5\sqrt3.$

Topics:Equation of a circle

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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 73