Find the equation of the circle passing through the point (- 1, 3) and having its centre at the intersection of the lines x - 2y = 4 and 2x + 5y + 1 = 0 ?
Here, we have to find the equation of the circle passing through the point (- 1, 3) and having its centre at the intersection of the lines x - 2y = 4 and 2x + 5y + 1 = 0 First let's find the point of intersection of the lines x - 2y = 4 and 2x + 5y + 1 = 0 By solving the equationsx - 2y = 4 and 2x + 5y + 1 = 0 we get x = 2 and y = - 1 So, the centre of the required circle is at (2, - 1) and let the radius be r units. As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x - h) + (y - k) = r Here, we have h = 2 and k = - 1 ⇒(x−2)2+(y+1)2=r2 .........(1) ∵ The required circle also passes through the point (- 1, 3) So, x = - 1 and y = 3 will satisfy the equation (1) ⇒(−1−2)2+(3+1)2=r2 ⇒r2=25 So, the equation of the required circle is (x−2)2+(y+1)2=25 ⇒x2+y2−4x+2y−20=0 So, the equation of the required circle is x2+y2−4x+2y−20=0 Hence, option B is the correct answer.