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

Concept: Circle Equation Standard Form — $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the centre and $r$ the radius.

Step 1: Centre lies on $y = x - 1$, so let centre be $(h, h-1)$. Radius $r = 3$, so the equation is

Step 2: The circle passes through $(7,3)$. Substitute:

Step 3: Expand and simplify:

$(49 - 14h + h^2) + (16 - 8h + h^2) = 9$

$\Rightarrow 2h^2 - 22h + 65 = 9$

$\Rightarrow 2h^2 - 22h + 56 = 0$ …

The standard form of a circle’s equation is a powerful tool: it tells you everything about the circle — its centre and its radius — in one clean expression. When a problem gives you the radius and a condition on the centre, you’re essentially being asked to find which specific centre (among those satisfying the condition) also makes the circle pass through the given point.

Here, the centre lies on the line $y = x - 1$. That means if the centre is $(h, k)$, then $k = h - 1$. So the centre is not arbitrary; it’s completely determined by a single variable $h$. This reduces the problem to solving for $h$.

1. Write the circle equation with the centre condition

   Let the centre be $(h, k)$. Since the centre lies on $y = x - 1$, we have:

The radius is given as $3$, so $r = 3$. The equation of the circle is:

2. Plug in the given point $(7, 3)$

   The circle passes through $(7, 3)$, so these coordinates must satisfy the equation:

Simplify the second term:

So the equation becomes:

3. Expand and solve for $h$

   Expand both squares:

Combine like terms:

Subtract 9 from both sides:

Divide through by 2:

Factor the quadratic:

So $h = 4$ or $h = 7$.

4. Find the corresponding centres and write the equations

   For $h = 4$: $k = h - 1 = 3$. Centre is $(4, 3)$. Equation: …