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 $y=x-1$
$\therefore k=h-1$
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 $x^2+y^2-14x+12y+76=0$
and $(x-4{)}^2+(y-3{)}^2=9$ or $x^2+y^2-8x-6y+16=0$