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Find the equation of the circle whose centre is (2, – 5) and which passes through the point (3, 2). |
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Answer» The general form of the equation of a circle is: (x – h)2 + (y – k) 2 = r2 …..(1) Where, (h, k) is the centre of the circle. r = radius of the circle. We are given with, centre = (2, – 5) Or (h, k) = (2, – 5) Find the radius of circle: Since the circle passes through (3, 2), so it must satisfy the equation. Put x = 3 and y = 2 in (1) (3 – 2)2 + (2 + 5) 2 = r2 1 + 49 = r2 Or r2 = 50 Now, Equation of circle is: (x – 2)2 + (y + 5)2 = 50 Which is required equation. |
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