Q.Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16$.

Concept: Circle Equation Standard Form

The general equation is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the centre.

Steps

1. Centre $(h, k)$ lies on $4x + y = 16$, so $k = 16 - 4h$.

2. Distance from centre to each point equals the radius:

   $(h - 4)^2 + (k - 1)^2 = (h - 6)^2 + (k - 5)^2$.

3. Substitute $k = 16 - 4h$ and simplify:

   $(h - 4)^2 + (15 - 4h)^2 = (h - 6)^2 + (11 - 4h)^2$ …

We start with the standard form of a circle:

where $(h,k)$ is the centre and $r$ is the radius. The problem gives two points on the circle and a line that the centre must lie on. That gives us three conditions to find $h$, $k$, and $r$.

Why this approach works:

Instead of trying to guess the centre, we write equations that must be true for the given points. Each point gives one equation. The line condition gives a third equation. Three unknowns, three equations — solvable.

1. Point $(4,1)$ lies on the circle Substitute $x=4$, $y=1$:

2. Point $(6,5)$ lies on the circle Substitute $x=6$, $y=5$:

3. Centre lies on the line $4x+y=16$ So $4h + k = 16 \quad \text{(3)}$

Now, since both (1) and (2) equal $r^2$, we can set them equal to each other:

Expand both sides:

• Left: $(16 - 8h + h^2) + (1 - 2k + k^2) = h^2 + k^2 - 8h - 2k + 17$
• Right: $(36 - 12h + h^2) + (25 - 10k + k^2) = h^2 + k^2 - 12h - 10k + 61$

Cancel $h^2 + k^2$ from both sides:

Bring terms together:

Divide by 4:

Now we have two linear equations in $h$ and $k$:

• From (3): $4h + k = 16$
• From (4): $h + 2k = 11$

Solve these simultaneously. From (4): $h = 11 - 2k$. Substitute into (3):

Then $h = 11 - 2(4) = 3$. …