Question

The equation of the circle which passes through the points (5, 1) and (7, -3) and the centre lies on the straight line 4x + 3y + 1 = 0, is ______ 

विकल्प

• x² + y² + 4x - 10y + 12 = 0

• x² + y² - 4x - 10y + 12 = 0

• x² + y² - 4x + 6y - 12 = 0

• x² + y² - 4x + 6y + 12 = 0

Answer 1

The equation of the circle which passes through the points (5, 1) and (7, -3) and the centre lies on the straight line 4x + 3y + 1 = 0, is x² + y² - 4x + 6y - 12 = 0.

Explanation:

Let centre be (h, k). Then,

`sqrt((h - 5)^2 + (k - 1)^2) = sqrt((h - 7)^2 + (k + 3)^2)`

⇒ -10h + 25 - 2k + 1 = -14h + 49 + 6k + 9

⇒ 4h - 8k - 32 = 0

⇒ h - 2k - 8 = 0 .... (i)

Since, centre lies on the given line.

∴ 4h + 3k + 1 = 0 .............(ii)

Solving (i) and (ii), we get

(h, k) = (2, -3)

∴ centre is (2, -3) and

radius = `sqrt((2 - 5)^2 + (-3 - 1)^2) = 5`

∴ the required equation of the circle is

`(x - 2)^2 + (y + 3)^2 = (5)^2`

⇒ x² + y² - 4x + 6y - 12 = 0