Question

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Answer 1

We know that the intersection point of the diameter gives the centre of the circle.

Given equations of diameters are

2x – 3y = 5   ......(i)

3x – 4y = 7   ......(ii)

From equation (i) we have x = `(5 + 3y)/2`  ......(iii)

Putting the value of x in eq. (ii) we have

`3((5 + 3y)/2) - 4y` = 7

⇒ 15 + 9y – 8y = 14

⇒ y = 14 – 15

⇒ y = – 1

Now from equation (iii) we have

x = `(5 + 3(-1))/2`

⇒ x = `(5 - 3)/2`

⇒ x = 1

So, the centre of the circle = (1, – 1)

Given that area of the circle = 154

⇒ `pir^2` = 154

⇒ `2/7 xx r^2` = 154

⇒ `r^2 = 154 xx 7/22`

⇒ `r^2 = 7 xx 7`

⇒ r = 7

So, the equation of the circle is (x – 1)² + (y + 1)² = (7)²

⇒ x² + 1 – 2x + y² + 1 + 2y = 49

⇒ x² + y² – 2x + 2y = 47

Hence, the required equation of the circle is x² + y² – 2x + 2y = 47.