Q.22. Circle, Parabola, Tangents, Circumcentre

Question 22.
Consider the circle x^2 + y^2 = 9 and the parabola y^2 = 8x. They intersect at P and Q in the first and fourth quadrant respectively .Tangents to the circle at P and Q intersect the x axis at R and tangents to the parabola at P and Q intersect the x axis at S. The radius of the circumcircle of the triangle PRS is____?

Answer 22.
Solving the equations
x^2 + y^2 = 9 ... (1) and
y^2 = 8x ... (2),
x^2 + 8x - 9 = 0
=> (x - 1)(x + 9) = 0
=> x = 1 [x = -9 is not possible]
and y = ± 2√2
=> (1, 2√2) and (1, - 2√2) are the points of intersection.

Eqn. of tangent at (1, 2√2) to the circle is
x + (2√2)y - 9 = 0
It intersects x-axis at R(9, 0)

Eqn. of tangent at (1, 2√2) to the parabola is
2√2 y = 4(x + 1)
It intersects x-axis at S(-1, 0)

We have to find circumradius of triangle PRS, where
P = (1, 2√2), R = (9, 0) and S = (-1, 0)
Perpendicular bisector of RS is x = 4
Perpendicular bisector of PS is given by
(x - 1)^2 + (y - 2√2)^2 = (x + 1)^2 + y^2
=> 4x + 4√2y = 8
=> x + √2y = 2
Solving with x = 4, y = - √2
=> circumcentre is (4, - √2)
Circumradius is its distance from (-1, 0)
= √[(4 + 1)^2 + (√2)^2]
= 3√3.


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