JEE Main 2019 — 10 January, Morning Slot. Previous Year Question.
Problem
If a circle $C$ passing through the point $(4,0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1,-1)$, then the radius of $C$ is:
Key insight. When two circles touch externally, their centres and the point of tangency all lie on one straight line — with the tangency point sitting between the two centres. This means circle $C$’s centre lies at a known distance $r$ (its own radius) from $(1,-1)$, along the exact direction away from the known circle’s centre — turning the problem into finding a single unknown, $r$.
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Approach
Find the centre and radius of the given circle. Determine the direction from its centre through the tangency point $(1,-1)$, and express $C$’s centre as a point along this same direction, at distance $r$ beyond $(1,-1)$. Use the fact that $(4,0)$ also lies on circle $C$ (so it’s at distance $r$ from $C$’s centre too) to set up and solve one equation in $r$.
Solution
Step 1 — Find the given circle’s centre and radius
For $x^2+y^2+4x-6y-12=0$: centre $=(-2,3)$, radius $=\sqrt{4+9+12}=\sqrt{25}=5$.
Step 2 — Confirm the tangency point lies at the known circle’s radius
Distance from $(-2,3)$ to $(1,-1)$: $\sqrt{(1+2)^2+(-1-3)^2} = \sqrt{9+16} = \sqrt{25}=5$ ✓ (matches the known radius, confirming $(1,-1)$ is indeed on this circle).
Step 3 — Find the unit direction from the known centre through the tangency point
$$\text{Direction} = (1-(-2),\ -1-3) = (3,-4), \qquad \text{unit vector} = \left(\frac{3}{5},-\frac{4}{5}\right)$$
Step 4 — Express C’s centre in terms of its radius r
Since the circles touch externally, $C$’s centre lies further along this same direction, at distance $r$ beyond the tangency point:
$$\text{Centre of }C = (1,-1) + r\left(\frac{3}{5},-\frac{4}{5}\right) = \left(1+\frac{3r}{5},\ -1-\frac{4r}{5}\right)$$
Step 5 — Use the fact that (4,0) is also on circle C
The distance from $C$’s centre to $(4,0)$ must also equal $r$:
$$\left(1+\frac{3r}{5}-4\right)^2 + \left(-1-\frac{4r}{5}-0\right)^2 = r^2$$
$$\left(\frac{3r}{5}-3\right)^2 + \left(-1-\frac{4r}{5}\right)^2 = r^2$$
Step 6 — Expand both squared terms
$$\left(\frac{3r}{5}-3\right)^2 = 9 – \frac{18r}{5} + \frac{9r^2}{25}$$
$$\left(1+\frac{4r}{5}\right)^2 = 1 + \frac{8r}{5} + \frac{16r^2}{25}$$
Step 7 — Combine and simplify
$$9-\frac{18r}{5}+\frac{9r^2}{25} + 1+\frac{8r}{5}+\frac{16r^2}{25} = r^2$$
$$10 – \frac{10r}{5} + \frac{25r^2}{25} = r^2 \implies 10-2r+r^2 = r^2$$
Step 8 — Solve for r
$$10-2r = 0 \implies r=5$$
Answer
$$5$$
Common mistakes
- Placing $C$’s centre on the wrong side of the tangency point. For external tangency, both centres lie on the same side of the tangency point along the line joining them (with the point between the two centres) — placing $C$’s centre in the opposite direction would model internal tangency instead, giving a wrong equation.
- Forgetting to verify the tangency point actually lies on the given circle first. This quick check (Step 2) confirms the problem’s setup is consistent before building on it.
Practise next
- A circle $D$ passes through $(7,5)$ and touches the circle $x^2+y^2-2x-4y=20$ externally at $(4,6)$. Find the radius of $D$, using the same line-of-centres approach.
Show answer
$5$. The given circle has centre $C_1=(1,2)$ and radius $5$, and $(4,6)$ does lie on it: $9+16=25$.
External tangency puts $C_1$, the point of contact and the centre of $D$ on one line, with the contact point between them. The unit vector from $C_1$ towards $(4,6)$ is $\tfrac15(3,4)$, so the centre of $D$ is $(4,6)+k\cdot\tfrac15(3,4)$ for its radius $k$.
Requiring $D$ to pass through $(7,5)$ gives $k=5$, so the centre is $(7,10)$ — and indeed $|C_1C_2|=10=5+5$, confirming external contact.
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