Question

The equation of the circle which passes through the points

(2, 3) and (4, 5) and the centre lies on the straight line $y - 4 x + 3 = 0$, is.

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

Answer 1

Answer: (B)

$x ^ { 2 } + y ^ { 2 } - 4 x - 10 y + 25 = 0$

Answer 2

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 $k - 4 h + 3 = 0$ ….(ii)

From (i), we get $- 4 h - 6 k + 8 h + 10 k = 16 + 25 - 4 - 9$

or $4 h + 4 k - 28 = 0$ 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 $x ^ { 2 } + y ^ { 2 } - 4 x - 10 y + 25 = 0$

passes through (2, 3) and (4, 5).