Question

The equation of the circle which passes through the points \((2,3)\) and \((4,5)\) and whose centre lies on a straight line \(4x-y-3 = 0\), is

• A. \((x-1)^2+(y-6)^2 = 10\)
• B. \((x-3)^2+(y-4)^2 = 2\)
• C. \(x^2+(y-7)^2 = 20\)
• D. \((x-2)^2+(y-5)^2 = 4\)

Hint:

Test each option: the centre must lie on 4x - y - 3 = 0 and be equidistant from both points.

Answer 1

The Correct Option is D

Step 1: Use the general condition:
Let the centre be $(h, 4h - 3)$ since it lies on $y = 4x - 3$.

Step 2: Equal distances:
$(h-2)^2 + (4h-6)^2 = (h-4)^2 + (4h-8)^2$.
Left: $h^2 - 4h + 4 + 16h^2 - 48h + 36 = 17h^2 - 52h + 40$.
Right: $h^2 - 8h + 16 + 16h^2 - 64h + 64 = 17h^2 - 72h + 80$.
Equating gives $20h = 40$, so $h = 2$, $k = 5$.

Step 3: Radius:
$r^2 = (2-2)^2 + (5-3)^2 = 4$. The equation is $(x-2)^2 + (y-5)^2 = 4$.

Final Answer:
Option (D) is the circle. \[ \boxed{(x-2)^2+(y-5)^2=4} \]