Question

If the tangent at the point P on the circle x² + y2 + 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​

Answer 1

The equation of the circle is,

x² + y² + 6x + 6y = 2

x² + 6x + 9 + y² + 6y + 9 = 2 + 18

(x + 3)² + (y + 3)² = 20

So the circle is centred at the point O(-3, -3) (say) with radius √20 = 2√5 units.

If the tangent of the circle at the point P meets a straight line of the equation 5x - 2y + 6 = 0 along with the y - axis at Q, the x coordinate of Q will be 0. Hence,

5(0) - 2y + 6 = 0

=> y = 3

So the point Q has the coordinate (0, 3).

Now, to find the length of PQ, first we have to find the length of OQ and OP.

Well, OP is a radius, so OP = 2√5 units.

And,

OQ = √((- 3 - 0)² + (- 3 - 3)²)

OQ = √(9 + 36)

OQ = 3√5 units.

Here, ∆OPQ is a right triangle, so we have, by Pythagoras' Theorem,

OQ = √((OQ)² - (OP)²)

OQ = √((3√5)² - (2√5)²)

OQ = √(45 - 20)

OQ = 5 units

Answer 2

Answer:

The equation of circle is

$x ^{2} + y ^{2} + 6x + 6y = 2$

and the equation of straight line is 5x-2y+6=0

Then taking x intercept:

5(0)-2y+6=0 then on solving we get y=3 and x=0

Then substituting x and y values in the length of tangent formula we get:

L=root of equation of circle

$\sqrt{x ^{2} + y {}^{2} + 6x + 6y - 2 }$

on solving root of 3^2+6×3-2 we get root25 =5

hence length of tangent is 5units