If tangent at point P on circle {x^2} + {y^2} + 6x + 6y = 2 meets straight line 5x - 2y + 6 = 0 at a

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10-1.Circle and System of Circles

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If the tangent at the point $P$ on the circle ${x^2} + {y^2} + 6x + 6y = 2$ meets the straight line $5x - 2y + 6 = 0$ at a point $Q$ on the $y$- axis, then the length of $PQ$ is

A

$4$

B

$2\sqrt 5 $

C

$5$

D

$3\sqrt 5 $

(IIT-2002)

Solution

(c) Let $P = ({x_1},{y_1})$ The tangent at $P$ is

$x{x_1} + y{y_1} + 3(x + {x_1}) + 3(y + {y_1}) – 2 = 0$…..(i)

Co-ordinates of $Q$ satisfy (i), $5x – 2y + 6 = 0,\,x = 0.$

So, $3{x_1} + 6{y_1} + 7 = 0$ and $Q = (0,\,3)$

$\therefore $ $P{Q^2} = x_1^2 + {({y_1} – 3)^2} = x_1^2 + y_1^2 – 6{y_1} + 9$

$ = 11 – 6{x_1} – 12{y_1}$, $(\because \,x_1^2 + y_1^2 + 6{x_1} + 6{y_1} – 2 = 0)$

$ = 11 – 2(3{x_1} + 6{y_1}) = 11 – 2( – 7) = 25$. So, $PQ =5$.

Standard 11

Mathematics

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