Question

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, is

• A. $x^2+y^2+4x-10y+25=0$
• B. $x^2+y^2-4x-10y+25=0$
• C. $x^2+y^2-4x-10y+16=0$
• D. $x^2+y^2-14y+8=0$

Answer 1

The correct option is B $x^2+y^2-4x-10y+25=0$

First find the centre. Let centre be (h, k), then
$\sqrt{(h-2{)}^2+(k-3{)}^2}=\sqrt{(h-4{)}^2+(k-5{)}^2}$ ….(i)
and k - 4h + 3 = 0 ….(ii)
From (i), we get -4h - 6k + 8h + 10k = 16 + 25 - 4 - 9 or 4h + 4k - 28 = 0 or h + k - 7 = 0 ....(iii)
From (iii) and (ii), we get (h, k) as (2, 5). Hence centre is (2, 5) and radius is 2. Now find the equation of circle.

Trick : Obviously, circle x^{2} + y^{2} - 4x - 10y + 25 = 0 passes through (2, 3) and (4, 5).