Consider the circle x2+y2=9 and the parabola y2=8 x. They intersect at P and Q in the first and the fourth quadrants, 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

Question

Consider the circle x2+y2=9 and the parabola y2=8 x. They intersect at P and Q in the first and the fourth quadrants, 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 (A) 5 (B) 3 √3 (C) 3 √2 (D) 2 √3

Tardigrade Admin · Accepted Answer

The circumcircle of triangle PRS is (x+1)(x-9)+y2+λ y=0 It will pass through the point (1,2 √2). Then, λ=2 √2 The equation of circumcircle is x2+y2-8 x+2 √2 y-9=0 Hence, its radius is 3 √3.

$5$

$33$

$32$

$23$

The circumcircle of $△ PRS$
is $(x + 1) (x - 9) + y^{2} + λ y = 0$
It will pass through the point $(1, 22)$ . Then,
$λ = 22$
The equation of circumcircle is
$x^{2} + y^{2} - 8 x + 22 y - 9 = 0$
Hence, its radius is $33$ .

5

33​

32​

23​

The circumcircle of △PRS is (x+1)(x−9)+y2+λy=0 It will pass through the point (1,22​). Then, λ=22​ The equation of circumcircle is x2+y2−8x+22​y−9=0 Hence, its radius is 33​.

$5$

$33$

$32$

$23$

The circumcircle of $△ PRS$
is $(x + 1) (x - 9) + y^{2} + λ y = 0$
It will pass through the point $(1, 22)$ . Then,
$λ = 22$
The equation of circumcircle is
$x^{2} + y^{2} - 8 x + 22 y - 9 = 0$
Hence, its radius is $33$ .