Find the equation of the circle which passes through the points (3, 7)

1.	Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1.
Answer» Given: The points (3, 7), (5, 5) The line x – 4y = 1…. (1) By using the standard form of the equation of the circle: x² + y² + 2ax + 2by + c = 0 ….. (2) Let us substitute the centre (-a, -b) in equation (1) we get, (- a) – 4(- b) = 1 -a + 4b = 1 a – 4b + 1 = 0…… (3) Substitute the points (3, 7) in equation (2), we get 3² + 7² + 2a(3) + 2b(7) + c = 0 9 + 49 + 6a + 14b + c = 0 6a + 14b + c + 58 = 0….. (4) Substitute the points (5, 5) in equation (2), we get 5² + 5² + 2a(5) + 2b(5) + c = 0 25 + 25 + 10a + 10b + c = 0 10a + 10b + c + 50 = 0….. (5) By simplifying equations (3), (4) and (5) we get, a = 3, b = 1, c = – 90 Now, by substituting the values of a, b, c in equation (2), we get x² + y² + 2(3)x + 2(1)y – 90 = 0 x² + y² + 6x + 2y – 90 = 0 ∴ The equation of the circle is x² + y² + 6x + 2y – 90 = 0.

Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1.

Answer»

Given:

The points (3, 7), (5, 5)

The line x – 4y = 1…. (1)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (2)

Let us substitute the centre (-a, -b) in equation (1) we get,

(- a) – 4(- b) = 1

-a + 4b = 1

a – 4b + 1 = 0…… (3)

Substitute the points (3, 7) in equation (2), we get

3² + 7² + 2a(3) + 2b(7) + c = 0

9 + 49 + 6a + 14b + c = 0

6a + 14b + c + 58 = 0….. (4)

Substitute the points (5, 5) in equation (2), we get

5² + 5² + 2a(5) + 2b(5) + c = 0

25 + 25 + 10a + 10b + c = 0

10a + 10b + c + 50 = 0….. (5)

By simplifying equations (3), (4) and (5) we get,

a = 3, b = 1, c = – 90

Now, by substituting the values of a, b, c in equation (2), we get

x² + y² + 2(3)x + 2(1)y – 90 = 0

x² + y² + 6x + 2y – 90 = 0

∴ The equation of the circle is x² + y² + 6x + 2y – 90 = 0.