Question

Find the equation of a circle passing through the point $(7,3)$ having radius $3$ units and whose centre lies on the line $y=x-1$.

Answer 1

Let $(a,b)$ be the centre of the required circle.

Given that the radius of the circle is 3 units.

Thus, the equation of the circle is
$(x-a{)}^2+(y-b{)}^2=3^2\dots \left(1\right)$

Also given that the point (7, 3) is passing through the required circle.

Hence we have,
$(7-a{)}^2+(3-b{)}^2=3^2\Rightarrow 49+a^2-14a+9+b^2-6b=9\Rightarrow a^2+b^2-14a-6b+49=0\dots \left(2\right)$

Given that the centre of the required circle lies on the line $y=x-1$

That is we have, $b=a-1$

Substituting the value $b$ as $a-1$ in equation (2),

$a^2+(a-1{)}^2-14a-6(a-1)+49=0\Rightarrow 2a^2-22a+56=0\Rightarrow a^2-11+28=0\Rightarrow a^2-7a-4a+28=0\Rightarrow (a-7\left)\right(a-4)=0\Rightarrow a-7=0ora-4=0\Rightarrow a=7orb=4$

Consider the root $a=4$,

Since, $b=a-1$, we have, $b=3$

Thus, from equation (1), we have,

$(x-4{)}^2+(y-3{)}^2=3^2$

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

If $a=7$, then $b=7-1=6$

Thus the equation becomes,

$(x-7{)}^2+(y-6{)}^2=3^2$

$\Rightarrow x^2+y^2-14x-12y+76=0$