Question

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

• A. $x^2 + y^2 + 8x - 6y +16 = 0$
• B. $x^2 + y^2 - 8x + 6y +16 = 0$
• C. $x^2 + y^2 - 8x - 6y -16 = 0$
• D. $x^2 + y^2 - 8x - 6y +16 = 0$

Answer 1

Answer: (D)

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

Answer 2

Let the centre of the circle be $(h, k).$
Since the centre lies on the line $y = x - 1$
$\therefore k-h-1\,...\left(i\right)$
Since the circle passes through the point $(7, 3)$, therefore, the distance of the centre from this point is the radius of the circle.
$\therefore 3=\sqrt{\left(h-7\right)^{2}+\left(k-3\right)^{2}}$
$\Rightarrow 3=\sqrt{\left(h-7\right)^{2}+\left(h-1-3\right)^{2}}\,\left[using \left(i\right)\right]$
$\Rightarrow h = 7$ or $h = 4$
For $h = 7$, we get $k = 6$ and for $h = 4$, we get $k = 3$
Hence, the circles which satisfy the given conditions are:
$\left(x-7\right)^{2}+\left(y-6\right)^{2}=9$
or $x^{2}+y^{2}-14x+12y+76=0$
and $\left(x-4\right)^{2}+\left(y-3\right)^{2}=9$ or
$x^{2}+y^{2}-8x-6y+16=0$