Q.Find the centre and radius of the circle $x^2 + y^2 - 8x + 10y - 12 = 0$.

Concept: Standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$ has centre $(h, k)$ and radius $r$. We complete the square in both variables.

Step 1: Group $x$ and $y$ terms separately:

Step 2: Complete the square for $x$: take half the coefficient of $x$, square it: $\left(\frac{-8}{2}\right)^2 = 16$. Add and subtract:

Step 3: Complete the square for $y$: $\left(\frac{10}{2}\right)^2 = 25$. Add and subtract:

$(y^2 + 10y + 25) - 25$ …

Why completing the square works

The standard form of a circle's equation $(x - h)^2 + (y - k)^2 = r^2$ immediately tells us the centre $(h, k)$ and radius $r$. When we're given the general form $x^2 + y^2 + Dx + Ey + F = 0$, the squared terms are "hidden" inside expanded binomials. Completing the square reverses this expansion, collecting the $x$-terms into $(x - h)^2$ and the $y$-terms into $(y - k)^2$.

The technique hinges on a simple algebraic identity: to turn $x^2 + bx$ into a perfect square, add and subtract $\left(\frac{b}{2}\right)^2$, giving $\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2$.

Step-by-step solution

1. Group the $x$-terms and $y$-terms separately

Starting with:

Rearrange to collect like variables:

I've moved the constant $-12$ to the right side.

2. Complete the square for the $x$-terms

For $x^2 - 8x$, take half the coefficient of $x$: $\frac{-8}{2} = -4$, then square it: $(-4)^2 = 16$.

Add and subtract $16$:

3. Complete the square for the $y$-terms

For $y^2 + 10y$, take half the coefficient of $y$: $\frac{10}{2} = 5$, then square it: $(5)^2 = 25$.

Add and subtract $25$:

4. Substitute back and simplify

Replace the grouped terms with their completed-square forms: …