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Find the equation of the circle passing through the point ( - 1, - 3) and having its centre at the point of intersection of the lines x – 2y = 4 and 2x + 5y + 1 = 0.

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The intersection of the lines: x – 2y = 4 and 2x + 5y + 1 = 0. is (2, - 1) 

∴ This problem is same as solving a circle equation with centre and point on the circle given.

The general form of the equation of a circle is: 

(x - h)2 + (y - k)2 = r2 

Where, (h, k) is the centre of the circle. 

r is the radius of the circle. 

In this question we know that (h, k) = (2, - 1), so for determining the equation of the circle we need to determine the radius of the circle.

Since the circle passes through ( - 1, - 3), that pair of values for x and y must satisfy the equation and we have: 

⇒ ( - 1 - 2)2 + ( - 3 - ( - 1))2 = r2 

⇒ ( - 3)2 + ( - 2)2 = r2

⇒ r2 = 9 + 4 = 13

∴ r2 = 13 

⇒ Equation of circle is:

(x - 2)2 + (y - ( - 1))2 = 13 

⇒ (x - 2)2 + (y + 1)2 = 13 

Ans:(x - 2)2 + (y + 1)2 = 13

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