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Solved Examples on Circle

Illustration 1: Tangents drawn from the point P (1, 8) to the circle x2 + y2 – 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)

Solution: The given equation is x2 + y2 – 6x – 4y – 11 = 0

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, x2 + y2 – 4x – 10y + 19 = 0

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 (x1, y1) and touching the straight line L is given by

(x-x1)2 + (y-y1)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

x2 + y2 – 4y + 4 + 5x = 0

or x2 + y2 + 5x – 4y + 4 = 0

For x-intercept we put y = 0.

Hence, x2 + 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...

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