Question

A circle has a radius of 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through point (7, 3).

Options

• x² + y² – 8x – 6y – 16 = 0

• x² + y² – 8x – 6y + 16 = 0

• x² + y² + 8x + 6y + 16 = 0

• x² + y² + 8x – 6y + 16 = 0

Answer 1

x² + y² – 8x – 6y + 16 = 0

Explanation:

`(x + h)^2 + (y - k)^2 = r^2`

Here `(h, k) ->` Centre

`k = h - 1`  (Given) ......(1)

And `(x, y)` point on circle

So given `(x, y)` = (7, 3)

∴ `(7 - h)^2 + (3 - h + 1)^2 = 3^2`

`(49 + h^2 - 14h) + (16 + h^2 - 8h)` = 9

⇒ `2h^2 - 22h + 56` = 0

⇒ `h^2` = 7 and 4

∴ `h` = 7 and 4

Put the value and get value of k = 6 and 3

∴ eqⁿ of circle corresponding

`h, k (4, 3) -> = (x - h)^2 + (y - k)^2 = r^2`

= `(x - 4)^2 + (y - 3)^2 = 3^2`

⇒ `x^2 + y^2 - 8x - 6y + 16` = 0