Question

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

Answer 1

Let the equation of the circle be

(x – h)² + (y – k)² = r²   ......(i)

If the circle passes through (2, 3) and (4, 5) then

(2 – h)² + (3 – k)² = r²  ......(ii)

And (4 – h)² + (5 – k)² = r²   ......(iii)

Subtracting equation (iii) from equation (ii) we have

(2 – h)² – (4 – h)² + (3 – k)² – (5 – k)² = 0

⇒ 4 + h² – 4h – 16 – h² + 8h + 9 + k² – 6k – 25 – k² + 10k = 0

⇒ 4h + 4k – 28 = 0

⇒ h + k = 7   ......(iv)

Since, the centre (h, k) lies on the line y – 4x + 3 = 0

Then k – 4h + 3 = 0

⇒ k = 4h – 3

Putting the value of k in equation (iv) we get

h + 4h – 3 = 7

⇒ 5h = 10

⇒ h = 2

From (iv) we get k = 5

Putting the value of h and k in equation (ii) we have

(2 – 2)² + (3 – 5)² = r²

⇒ r² = 4

So, the equation of the circle is (x – 2)² + (y – 5)² = 4

⇒ x² + 4 – 4x + y² + 25 – 10y = 4

⇒ x² + y² – 4x – 10y + 25 = 0

Hence, the required equation is x² + y² – 4x – 10y + 25 = 0.