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. [NCERT EXEMPLAR]

Answer 1

The general equation of the circle is x² + y² + 2gx + 2fy + c = 0 where the centre of the circle is (−g, −f)
Now, it is passing through (2, 3)
∴ 13 + 4g + 6f + c = 0 .....(1)
Also, it is passing through (4, 5)
∴ 41 + 8g + 10f + c = 0 .....(2)
⇒g=−a2
Now, the centre lies on the straight line y − 4x + 3 = 0
∴ −f + 4g + 3 = 0 .....(3)
⇒g=−a2
Solving (1), (2) and (3), we get
g = −2, f = −5 and c = 25
The equation of the circle is given by x² + y² − 4x − 10y + 25 = 0