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**Illustration 1: Tangents drawn from the point P (1, 8) to the circle x ^{2} + y^{2} – 6x – 4y – 11 = 0 touch the circle at the point A and B. What is the equation of the circum circle of the triangle PAB? (2009)**

Now, when we draw this circle and draw tangents from the given point i.e. P(1, 8) to the circle, we get

P(1, 8) and O(3, 2) as the end points of the diameter.

Therefore, we have (x-1) (x-3) + (y-8) (y-2) = 0

This gives, x

**Illustration 2: The circle passing through the point (-1, 0) and touching the y-axis at (0, 2) also passes through the point: (2011)****1. (-3/2, 0)****2. (-5/2, 2)****3. (-3/2, 5/2)****4. (-4, 0)****Solution:** Equation of a circle passing through a point (x_{1}, y_{1}) and touching the straight line L is given by (x-x_{1})^{2} + (y-y_{1})^{2} + μL = 0

Hence, the equation of circle passing through (0, 2) and touching x = 0 is

(x-0)^{2} + (y-2)^{2} + μx = 0 ….. (1)

Also, it passes through (-1, 0)

Hence, 1 + 4 – μ = 0

This gives the value of μ as 5.

Therefore, equation (1) becomes

x^{2} + y^{2} – 4y + 4 + 5x = 0

or x^{2} + y^{2} + 5x – 4y + 4 = 0

For x-intercept we put y = 0.

Hence, x^{2} + 5x + 4 = 0

(x+1)(x+4) = 0

So, this gives x = -1 and x = -4.

**Illustration 3: Find the equation of circle touching the line 2x+3y+1 = 0 at the point (1, -1) and is orthogonal to the circle which has the line segment having end points (0, -1) and (-2, 3) as the diameter. (2004)****Solution:** The equation of the circle having tangent 2x + 3y + 1 = 0 at the point (1, -1) is

(x - 1)^{2 }+ (y + 1)^{2} + μ(2x + 3y + 1) = 0

x^{2} + y^{2} + 2x(μ -1) + y(3μ + 2) + (μ+2) = 0 ….. (1)

This is orthogonal to the circle having end points of diameter as (0, -1) and (-2, 3)

Hence, x(x+2) + (y+1)(y-3) = 0

Or x^{2 }+ y^{2} + 2x -2y -3 = 0

Therefore, [2(2μ – 2)/2]. 1 + [2(3μ + 2)/ 2]. (-1) = μ + 2 – 3

Hence, we get 2μ - 2 - 3μ – 2= μ -1

This gives, 2μ = -3

Hence, μ = -3/2

Therefore, from equation (1) we get the equation of circle as 2x^{2} + 2y^{2 }-10x -5y + 1 = 0.

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