A Circle With Centre (2, 3) and Radius 4 Intersects the Line x + y = 3 at the Points P and Q. If the Tangents at P and Q Intersect at the Point S(α, β), Then 4α − 7β Is Equal To

JEE Main 2023 — 29 January, Evening Shift. Previous Year Question.

Problem

A circle with centre $(2,3)$ and radius $4$ intersects the line $x+y=3$ at the points $P$ and $Q$. If the tangents at $P$ and $Q$ intersect at the point $S(\alpha,\beta)$, then $4\alpha-7\beta$ is equal to?

Key insight. The line joining $P$ and $Q$ is itself the chord of contact from the external point $S$ — writing the general chord-of-contact formula from $S(\alpha,\beta)$ and matching it against the given line $x+y=3$ (since they describe the same line) turns the geometry problem into simple coefficient comparison.

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Approach

Write the circle’s equation in general form. Write the chord-of-contact formula from an external point $S(\alpha,\beta)$ using this equation. Since $P$ and $Q$ (where the tangents from $S$ touch the circle) lie on the given line $x+y=3$, the chord of contact and the given line must be the same line — compare their coefficients to solve for $\alpha$ and $\beta$.

Solution

Step 1 — Write the circle’s equation

Centre $(2,3)$, radius $4$: $(x-2)^2+(y-3)^2=16$

$$x^2-4x+4+y^2-6y+9=16 \implies x^2+y^2-4x-6y-3=0$$

Step 2 — Write the chord of contact from S(α, β)

For a circle $x^2+y^2+2gx+2fy+c=0$, the chord of contact from an external point $(\alpha,\beta)$ is:

$$x\alpha+y\beta+g(x+\alpha)+f(y+\beta)+c=0$$

Here $2g=-4\implies g=-2$; $2f=-6\implies f=-3$; $c=-3$.

$$x\alpha+y\beta-2(x+\alpha)-3(y+\beta)-3=0$$

Step 3 — Simplify

$$x(\alpha-2)+y(\beta-3) -2\alpha-3\beta-3 = 0$$

Step 4 — Compare with the given line x + y − 3 = 0

Since this chord of contact is the same line as $x+y=3$, i.e. $x+y-3=0$, the coefficients must be proportional:

$$\frac{\alpha-2}{1} = \frac{\beta-3}{1} = \frac{-2\alpha-3\beta-3}{-3}$$

Step 5 — Use the first equality

$$\alpha-2=\beta-3 \implies \alpha=\beta-1 \quad \text{…(i)}$$

Step 6 — Use the first and third parts together

$$-3(\alpha-2) = -2\alpha-3\beta-3$$

$$-3\alpha+6 = -2\alpha-3\beta-3$$

$$-\alpha+3\beta = -9 \implies \alpha = 3\beta+9 \quad \text{…(ii)}$$

Step 7 — Solve (i) and (ii) together

$$\beta-1 = 3\beta+9 \implies -10=2\beta \implies \beta=-5$$

$$\alpha = \beta-1 = -5-1=-6$$

Step 8 — Compute 4α − 7β

$$4\alpha-7\beta = 4(-6)-7(-5) = -24+35 = 11$$

Answer

Common mistakes

• Using the wrong chord-of-contact formula (mixing it up with the tangent-at-a-point formula). The chord of contact is drawn from an external point where two tangents touch the circle — its formula, $x\alpha+y\beta+g(x+\alpha)+f(y+\beta)+c=0$, differs subtly from the tangent-at-a-point-on-the-circle formula, and using the wrong one gives a completely different (and inconsistent) line.
• Setting up the proportionality between coefficients incorrectly. Since two linear equations represent the same line exactly when their coefficients (of $x$, $y$, and the constant) are all in the same ratio, all three ratios should be checked or used consistently — using only two coefficients while ignoring the third can hide an arithmetic error.

Practise next

• A circle with centre $(1,2)$ and radius $3$ intersects the line $2x+y=5$ at points $P$ and $Q$. If the tangents at $P$ and $Q$ meet at $S(\alpha,\beta)$, find $3\alpha-2\beta$, using the same chord-of-contact matching approach.