Question

Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18

Answer 1

Given equations are 3x + y = 14   ......(i)

And 2x + 5y = 18   ......(ii)

From equation (i) we get y = 14 – 3x   ......(iii)

Putting the value of y in equation (ii) we get

⇒ 2x + 5(14 – 3x) = 18

⇒ 2x + 70 – 15x = 18

⇒ – 13x = – 70 + 18

⇒ – 13x = – 52

∴ x = 4 

From equation (iii) we get, y = 14 – 3 × 4 = 2

∴ Point of intersection is (4, 2)

Now, radius r = `sqrt((4 - 1)^2 + (2 + 2)^2)`

= `sqrt((3)^2 + (4)^2)`

= `sqrt(9 + 16)`

= 5

So, the equation of circle is (x – h)² + (y – k)² = r²

⇒ (x – 1)² + (+ 2)² = (5)²

⇒ x² – 2x + 1 + y² + 4y + 4 = 25

⇒ x² + y² – 2x + 4y – 20 = 0

Hence, the required equation is x² + y² – 2x + 4y – 20 = 0.