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If a point P has coordinates (0, -2) and Q is any point on the circle $x^{2} + y^{2} - 5 x - y + 5 = 0$ , then maximum value of ${(PQ)}^{2} = 14 + 5 \sqrt{3}$ . State whether it’s true or false.

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True

False

answer is A.

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Detailed Solution

p = (0, - 2) Q is any point on the circle x^{2} + y^{2} - 5 x - y + 5 = 0

Compare given equation with

x^{2} + y^{2} + 2 gx + 2 fy + C = 0

we get,

2 g = - 5, 2 f = - 1

g = \frac{- 5}{2}, f = \frac{- 1}{2}

Center (o) = (- g, - f) = (\frac{5}{2}, \frac{1}{2})

r = \sqrt{g^{2} + f^{2} - c} = \sqrt{{(\frac{5}{2})}^{2} + {(\frac{1}{2})}^{2} - 5} = \sqrt{\frac{25}{4} + \frac{1}{4} - 5} = \frac{\sqrt{6}}{2}

Radius of circle=(OQ)=

\frac{\sqrt{6}}{2}

By distance formula

d (OP) = \sqrt{{(0 - \frac{5}{2})}^{2} + {(- 2 - \frac{1}{2})}^{2}}

d (OP) = \sqrt{{(- \frac{5}{2})}^{2} + {(\frac{- 4 - 1}{2})}^{2}}

d (OP) = \sqrt{{(- \frac{5}{2})}^{2} + {(\frac{- 5}{2})}^{2}}

d (OP) = \sqrt{\frac{25}{4} + \frac{25}{4}}

d (OP) = \sqrt{\frac{50}{4}}

d (OP) = \frac{\sqrt{50}}{\sqrt{4}} = \frac{\sqrt{2 * 25}}{2} = \frac{5 \sqrt{2}}{2}

By diagram

PQ = OP + r

PQ = \frac{\sqrt{6}}{2} + \frac{5 \sqrt{2}}{2}

{(PQ)}^{2} = {(\frac{\sqrt{6}}{2} + \frac{5 \sqrt{2}}{2})}^{2} = 14 + 5 \sqrt{3}

Hence, the given statement is true.

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