Question

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

• A. \(x^2+y^2-4x-10y+25 = 0\)
• B. \(x^2+y^2-4x-10y-25 = 0\)
• C. \(x^2+y^2-4x+10y-25 = 0\)
• D. \(x^2+y^2+4x-10y+25 = 0\)

Hint:

The centre is equidistant from both points and lies on $y=4x-3$.

Answer 1

The Correct Option is A

Step 1: Use the perpendicular bisector
The midpoint of the chord is $(3,4)$ and the slope of the chord is $1$, so the bisector has slope $-1$: $y-4=-(x-3)$, i.e. $x+y=7$.

Step 2: Intersect with the given line
$y=4x-3$ and $x+y=7$ give $5x=10$, so $x=2$, $y=5$. Radius$^2=4$.
Expand: $x^2+y^2-4x-10y+25=0$. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A)}} \]