JEE Main 2023 — 13 April, Evening Shift. Previous Year Question.
Problem
Let the centre of a circle $C$ be $(\alpha,\beta)$ and its radius $r<8$. Let $3x+4y=24$ and $3x-4y=32$ be two tangents and $4x+3y=1$ be a normal to $C$. Then $(\alpha-\beta+r)$ is equal to:
Key insight. A normal to a circle at any point always passes straight through the centre — so the centre must satisfy the normal line’s equation directly, no distance formula needed there. Combined with the fact that a tangent line sits at distance exactly $r$ from the centre, that’s enough equations to pin down $\alpha$, $\beta$, and $r$ all at once.
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Approach
Use the fact that the normal line passes through the centre to get one linear equation in $\alpha,\beta$. Set the perpendicular distances from $(\alpha,\beta)$ to both tangent lines equal to each other (since both equal the radius $r$), which gives a second equation after resolving the resulting absolute-value case correctly. Solve the system, discarding whichever sign choice gives a radius that violates $r<8$.
Solution
Step 1 — Use the normal line to get one equation
Since a normal to a circle always passes through its centre:
$$4\alpha+3\beta=1 \quad \text{…(i)}$$
Step 2 — Set the two tangent distances equal
Distance from $(\alpha,\beta)$ to $3x+4y-24=0$ equals distance to $3x-4y-32=0$ (both equal $r$):
$$\frac{|3\alpha+4\beta-24|}{5} = \frac{|3\alpha-4\beta-32|}{5} \implies |3\alpha+4\beta-24| = |3\alpha-4\beta-32|$$
Step 3 — Try the case with matching signs
$$3\alpha+4\beta-24 = 3\alpha-4\beta-32$$
$$8\beta = -8 \implies \beta=-1$$
Step 4 — Solve for α using equation (i)
$$4\alpha+3(-1)=1 \implies 4\alpha=4 \implies \alpha=1$$
Step 5 — Find r using the tangent-distance formula
$$r = \frac{|3(1)+4(-1)-24|}{5} = \frac{|3-4-24|}{5} = \frac{25}{5} = 5$$
Since $r=5<8$, this satisfies the given condition.
Step 6 — Check the opposite-sign case (for completeness)
Trying $3\alpha+4\beta-24=-(3\alpha-4\beta-32)$ instead leads to $\alpha=\dfrac{28}{3}$ and, after substituting into equation (i), a radius that comes out greater than $8$ — violating the given condition $r<8$, so this branch is rejected.
Step 7 — Compute α − β + r
$$\alpha-\beta+r = 1-(-1)+5 = 1+1+5 = 7$$
Answer
$$7$$
Common mistakes
- Forgetting that an absolute-value equation $|A|=|B|$ splits into two cases, $A=B$ and $A=-B$. Skipping the second case can hide the fact that one branch must be rejected using the given constraint ($r<8$) — that rejection is itself part of the intended solution, not an error to avoid.
- Using the distance formula for the normal line too. A normal doesn’t sit at distance $r$ from the centre — it passes through the centre (distance $0$). Treating it like a tangent and applying the distance formula gives a wrong, extra constraint.
Practise next
- Let a circle have centre $(\alpha,\beta)$ and radius $r<2$, with tangents $4x+3y=20$ and $4x-3y=16$, and normal $3x-4y=1$. Find $\alpha+\beta+r$, using the same normal-through-centre and equal-tangent-distance approach.
Show answer
$\dfrac{91}{10}$. A normal passes through the centre, so $3\alpha-4\beta=1$. Each tangent is at distance $r$ from the centre:
$$\frac{|4\alpha+3\beta-20|}{5}=\frac{|4\alpha-3\beta-16|}{5}=r.$$
Working through the sign cases gives exactly two circles: centre $\left(\tfrac92,\tfrac{25}{8}\right)$ with $r=\tfrac{59}{40}$, and centre $\left(\tfrac{11}{9},\tfrac23\right)$ with $r=\tfrac{118}{45}\approx2.62$.
$r<2$ keeps only the first, so $\alpha+\beta+r=\tfrac92+\tfrac{25}{8}+\tfrac{59}{40}=\tfrac{91}{10}$. The radius bound is doing real work here — without it the answer is not unique.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.