Find the equation of the circle passing through the point (0,5) and (6

1.	Find the equation of the circle passing through the point (0,5) and (6, 1) and has its centre on the line 12x + 5y = 25.
Answer» Let the equation of the required circle is x² + y² + 2gx + 2fy + c = 0 given that this Equation passes through the points (0,5) and (6, 1) ∴ (0,5) 02 + 52 + 2(0)g + 2(5)f + C=0 10f + C + 25 = 0 (6, 1) 62 + 12 + 2.6g + 2.1f+C = 0 …. (1) 12g + 2f + C + 37 = 0 …. (2) Also the centre (-g, -f) lies on the line 12x + 5y – 25 = 0 ∴ -128 – 5f – 22 = 0 …. (3) Equation (2) – Equation (1) gives 12g – 8f + 12 = 0 …(4) Adding equations 3 and 4 we get -13f – 13 = 0 → f = \(\frac{13}{-13}\) - 1 ⇒ f = -1 From equation 4 12g = 8f + 12 = – 8 + 12 = 4 g = \(\frac{1}{3}\) 10 + C + 25 = 0 ⇒ c = -15 ∴ Required equation is x² + y² + 2\(\frac{1}{3}\)x + 2(-1)y -15 = 0 ⇒ 3x² + 3y²+ 2x – y – 45 = 0

Find the equation of the circle passing through the point (0,5) and (6, 1) and has its centre on the line 12x + 5y = 25.

Answer»

Let the equation of the required circle is

x² + y² + 2gx + 2fy + c = 0 given that this

Equation passes through the points (0,5) and (6, 1)

∴ (0,5) 02 + 52 + 2(0)g + 2(5)f + C=0

10f + C + 25 = 0

(6, 1) 62 + 12 + 2.6g + 2.1f+C = 0 …. (1)

12g + 2f + C + 37 = 0 …. (2)

Also the centre (-g, -f) lies on the line

12x + 5y – 25 = 0

∴ -128 – 5f – 22 = 0 …. (3)

Equation (2) – Equation (1) gives

12g – 8f + 12 = 0 …(4)

Adding equations 3 and 4 we get

-13f – 13 = 0 → f = \(\frac{13}{-13}\) - 1

⇒ f = -1

From equation 4

12g = 8f + 12 = – 8 + 12 = 4

g = \(\frac{1}{3}\)

10 + C + 25 = 0 ⇒ c = -15

∴ Required equation is x² + y² + 2\(\frac{1}{3}\)x + 2(-1)y -15 = 0

⇒ 3x² + 3y²+ 2x – y – 45 = 0

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