Circle Through A(2,−1), B(3,4), Center on (x−5)²+(y−1)²=13/2, Find r²

JEE Main 2022 — 26 June, Morning Shift. Previous Year Question.

Problem

Let $C$ be a circle passing through the points $A(2,-1)$ and $B(3,4)$. The line segment $AB$ is not a diameter of $C$. If $r$ is the radius of $C$ and its centre lies on the circle $(x-5)^2+(y-1)^2=\dfrac{13}{2}$, then $r^2$ is equal to:

Key insight. Writing the circle in general form and matching coefficients leads to a messy system. The cleaner route uses a pure geometric fact: the centre of any circle through $A$ and $B$ must lie on the perpendicular bisector of chord $AB$ — so intersecting that perpendicular bisector with the given constraint circle locates the centre directly.

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Approach

Find the perpendicular bisector of $AB$ — since the centre of any circle through $A$ and $B$ must lie on it. Intersect this line with the given constraint circle $(x-5)^2+(y-1)^2=\frac{13}{2}$ to get the centre’s possible coordinates, reject the one that would make $AB$ a diameter, then compute $r^2$ as the squared distance from the centre to $A$.

Solution

Step 1 — Find the midpoint and slope of AB

Midpoint of $AB$: $\left(\dfrac{2+3}{2},\dfrac{-1+4}{2}\right) = \left(\dfrac{5}{2},\dfrac{3}{2}\right)$

Slope of $AB$: $\dfrac{4-(-1)}{3-2} = 5$

Step 2 — Write the perpendicular bisector

Its slope is $-\dfrac{1}{5}$ (negative reciprocal), passing through $\left(\dfrac{5}{2},\dfrac{3}{2}\right)$:

$$y – \frac{3}{2} = -\frac{1}{5}\left(x-\frac{5}{2}\right) \implies y = -\frac{x}{5}+2$$

Step 3 — Substitute into the constraint circle

$$(x-5)^2 + \left(-\frac{x}{5}+2-1\right)^2 = \frac{13}{2} \implies (x-5)^2 + \left(\frac{5-x}{5}\right)^2 = \frac{13}{2}$$

Since $\left(\dfrac{5-x}{5}\right)^2 = \dfrac{(x-5)^2}{25}$:

$$(x-5)^2\left(1+\frac{1}{25}\right) = \frac{13}{2} \implies (x-5)^2\cdot\frac{26}{25} = \frac{13}{2}$$

Step 4 — Solve for x

$$(x-5)^2 = \frac{13}{2}\times\frac{25}{26} = \frac{25}{4} \implies x-5 = \pm\frac{5}{2}$$

$$x = \frac{15}{2} \text{ or } x = \frac{5}{2}$$

Step 5 — Reject the invalid centre

At $x=\dfrac{5}{2}$: $y=-\dfrac{5/2}{5}+2=\dfrac{3}{2}$ — this is exactly the midpoint of $AB$, which would make $AB$ a diameter. Since the problem states $AB$ is not a diameter, this root is rejected.

So the centre is at $x=\dfrac{15}{2}$, giving $y=-\dfrac{15/2}{5}+2=\dfrac{1}{2}$. Centre $=\left(\dfrac{15}{2},\dfrac{1}{2}\right)$.

Step 6 — Compute r²

$$r^2 = \left(\frac{15}{2}-2\right)^2+\left(\frac{1}{2}-(-1)\right)^2 = \left(\frac{11}{2}\right)^2+\left(\frac{3}{2}\right)^2 = \frac{121}{4}+\frac{9}{4} = \frac{130}{4} = \frac{65}{2}$$

Answer

Common mistakes

• Missing the rejection step. Both roots of the quadratic satisfy the constraint circle, but only one is a valid centre for a circle where $AB$ is not a diameter — checking which root makes the centre coincide with the midpoint of $AB$ is essential, not optional.
• Trying the general-form approach (comparing coefficients in $x^2+y^2+2gx+2fy+c=0$) first. This leads to a system that isn’t linear once the given circle constraint is substituted in, making it far messier than the perpendicular-bisector method.

Practise next

• A circle passes through $P(1,2)$ and $Q(5,4)$, with $PQ$ not a diameter, and its centre lies on $(x-3)^2+(y-2)^2=1$. Find $r^2$, using the same perpendicular-bisector approach.