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: Understanding the Concept:
The centre \((h, k)\) of the circle must satisfy the line \(4h - k - 3 = 0\) and must be equidistant from \((2,3)\) and \((4,5)\), that is, on the perpendicular bisector of the chord joining them.

Step 2: Perpendicular bisector:
Midpoint of the two points is \((3,4)\). The chord has slope \(\frac{5-3}{4-2} = 1\), so the bisector has slope \(-1\): \(y - 4 = -(x - 3)\), that is, \(x + y = 7\).

Step 3: Find the centre:
Solve \(4x - y = 3\) and \(x + y = 7\). Adding gives \(5x = 10\), so \(x = 2\), \(y = 5\). Centre is \((2, 5)\).
Radius squared \(= (2-2)^2 + (3-5)^2 = 4\).

Step 4: Result:
The circle is \((x - 2)^2 + (y - 5)^2 = 4\).

Step 5: Why the other options are wrong.
Option (A) has centre \((1,6)\): \(4 - 6 - 3 \neq 0\). Option (B) has centre \((3,4)\): \(12 - 4 - 3 = 5 \neq 0\). Option (C) has centre \((0,7)\): \(0 - 7 - 3 \neq 0\).

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