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  4. A circle has radius 3 units and its centre lies on the line y=x-1. If it passes through the point (7,3), its equation is x2+y2+a x+b y+c=0. Here, a, b and c respectively are

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Q. A circle has radius 3 units and its centre lies on the line $y=x-1$. If it passes through the point $(7,3)$, its equation is $x^2+y^2+a x+b y+c=0$. Here, $a, b$ and $c$ respectively are

Conic Sections

Solution:

Let $(h, k)$ be the centre of the circle. Then, $k=h-1$. Therefore, the equation of the circle is given by
$(x-h)^2+[y-(h-1)]^2=9$
Given that, the circle passes through the point $(7,3)$ and hence we get
$(7-h)^2-(3-(h-1))^2=9$
or $(7-h)^2+(4-h)^2=9$
or $ h^2-11 h+28=0$
which given $(h-7)(h-4)=0$
$h=4$ or $h=7$
Therefore, the required equations of the circles are
$x^2+y^2-8 x-6 y+16=0$
or $ x^2+y^2-14 x-12 y+76=0$
Comparing above equations with
$x^2+y^2+a x+b y+c=0$
where, $ a--8, b--6, c-16$
and $ a=-14, b=-12, c=76$