Question

find the equation of the circle whose centre is the point (2,3)and which passes through the point of transaction of the lines x²+y²+4x-6y=12​

Answer 1

Answer:

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

∴ Required equation of the circle is

(x – 2)2 + (y + 3)^2 = (√109)^2

⇒ x^2 + y^2 – 4x + 6y – 96 = 0

Answer 2

Answer:

above answer is correct