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

Answer 1

Answer: (D)

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