Question

A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line, $y-4x+3=0$, then its radius is equal to :

• A. $\sqrt{2}$
• B. $\sqrt{5}$
• C. 2
• D. 1

Answer 1

Answer: (C)

Let $A(2,3)$ & $B(4,5)$ be the points through which circle is passing $\&$ radius lies on the line $y-4x+3=0$. Le centre be $(n,k)$, this point must satisfy the line $y-4x-+3=0$. Hence
$k-4n+3=0$
$k=4n-3.$
So, Centre coordinates be $0(n,4n-3)$
Now, $OA=OB$ (both one radius)
$A(2,3)B(4,5)$
$\Rightarrow OA=OB$
$\Rightarrow O{A}^2=O{B}^2$
$(n-2{)}^2+(4n-3-3{)}^2=(n-4{)}^2+(4n-3-5{)}^2$
$n^2+4-4n+16n^2+36-48n=n^2+16-8n+16n^2+64-64n$
$40-52n=80-72n$
$72n-52n=80-40$
$20n=40$
$n=2$
Hence $k=4n-3$
$k=8-3$
$k=5$
Centre coordinates $O(7,5)$ & $A(7,3)$
radius $=OA=(2-2{)}^2+(5-3{)}^2=2$
Hence = radius $=2$