Question

A circle has radius 3 units and its centre lies on the line $y = x - 1$. Then the equation of this circle if it passes through point (7, 3), is.

• A. $x ^ { 2 } + y ^ { 2 } - 8 x - 6 y + 16 = 0$
• B. $x ^ { 2 } + y ^ { 2 } + 8 x + 6 y + 16 = 0$
• C. $x ^ { 2 } + y ^ { 2 } - 8 x - 6 y - 16 = 0$
• D. None of these

Answer 1

Answer: (A)

$x ^ { 2 } + y ^ { 2 } - 8 x - 6 y + 16 = 0$

Answer 2

Let its centre be (h, k), then $h - k = 1$ ...(i)

Also radius $a = 3$

Equation is $( x - h ) ^ { 2 } + ( y - k ) ^ { 2 } = 9$

Also it passes through (7, 3)

i.e., $( 7 - h ) ^ { 2 } + ( 3 - k ) ^ { 2 } = 9$ ….(ii)

We get h and k from (i) and (ii) solving simultaneously as

(4, 3). Equation is $x ^ { 2 } + y ^ { 2 } - 8 x - 6 y + 16 = 0$.

Trick : Since the circle $x ^ { 2 } + y ^ { 2 } - 8 x - 6 y + 16 = 0$

satisfies the given conditions.