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)

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