# Find the equation of a circle with radius 5 whose centre lies on x axis and passes through the point (2,3)

2
by jiki

Log in to add a comment

by jiki

Log in to add a comment

There are 2 points in the x-axis that are 5 units from the point (2,3).

The points are (x,0).

d2= 25 = diffy2+diffx2=3 2+(2-x)2

x2-4x+13=25

x2-4x-12=0

(x - 6)*(x + 2) = 0

x = -2, x = 6

------------

Circle 1 (x+2)2+y2=5 2

Circle 2 (x-6)2+y2=5 2

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

If the center lies on x axis, y coordinate is zero. let the centre be (a, 0).

(x-a)² + (y-0)² = 5²

(2,3) lies on the circle. So, (2-a)²+3² = 5²

(2 - a)² = 25 - 9 = 16

2 - a = +4 or -4 => a = -2 or 6

Equation of the circle :

(x+2)² + y² = 5² with center at (-2, 0)

or (x-6)² + y² = 5² with center at (6,0)

(x-a)² + (y-0)² = 5²

(2,3) lies on the circle. So, (2-a)²+3² = 5²

(2 - a)² = 25 - 9 = 16

2 - a = +4 or -4 => a = -2 or 6

Equation of the circle :

(x+2)² + y² = 5² with center at (-2, 0)

or (x-6)² + y² = 5² with center at (6,0)