Question

Find the equation of the circle passing through the point of intersection of the lines X+3y
= 0 & 2x-7y=0 and whose centre is the point of intersection of lines X+y+1=0 and x-2y+4=0.​

Answer 1

The required circle is

(x + 4)² + (y - 3)² = 25

Step-by-step explanation:

The circle passes through the point of intersection of the lines,

x + 3y = 0 ..... (1)

2x - 7y = 0 ..... (2)

The point of intersection of (1) & (2) is (0, 0)

The centre of the circle is at the point of intersection of the lines,

x + y + 1 = 0 ..... (3)

x - 2y + 4 = 0 ..... (4)

From (3), we get

x = - y - 1

Putting x = - y - 1 in (4), we get

- y - 1 - 2y + 4 = 0

or, 3y = 3

i.e., y = 3

So x = - 3 - 1 = - 4

Thus (- 4, 3) is the centre of the circle.

Let us consider that, the required circle be

{x - (- 4)}² + (y - 3)² = c

or, (x + 4)² + (y - 3)² = c

or, x² + 8x + 16 + y² - 6y + 9 = c

or, x² + y² + 8x - 6y + 25 = c ..... (5)

Given that, the circle (5) passes through the point (0, 0). Then c = 25.

Hence the required circle is

(x + 4)² + (y - 3)² = 25,

whose centre is at the point (- 4, 3) and radius is 5 units.