Question

Equation of a circle passing through the intersection of line 3x-2y-1=0 and 4x-y-27=0 and its center is (2,-3) find equation of circle.

Answer 1

Step-by-step explanation:

Solving 3x – 2y = 1 and 4x + y = 27 Simultaneously, we get x = 5 and y = 7 ∴ The point of intersection of the lines is (5, 7)

Now we have to find the equation of a circle whose centre is (2, -3) and which passes through (5, 7)

Radius = √((5 - 2)^2 + (7 + 3)^2

= √(9 + 100) = √109

equation of circle=(x-h)^2+(y-k)^2=r^2

∴ Required equation of the circle is (x – 2)^2 + (y + 3)^2 = (√109)^2

⇒ x2 + y2 – 4x + 6y – 96 = 0