Power of a point with respect to a circle — tangent and secant

Question

A point $P(5, 6)$ lies outside the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ . Find the power of the point $P$ with respect to the circle. Also find the length of the tangent from $P$ to the circle.

(JEE Main 2023)

Solution — Step by Step

The power of a point $(x_1, y_1)$ with respect to the circle $S: x^2 + y^2 + 2gx + 2fy + c = 0$ is:

S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c

For $P(5, 6)$ and the given circle ( $g = -2, f = -3, c = -12$ ):

S_1 = 25 + 36 + 2(-2)(5) + 2(-3)(6) + (-12)

= 25 + 36 - 20 - 36 - 12 = \mathbf{-7}

Wait — the power is negative, which means $P$ is inside the circle, not outside.

Centre: $(-g, -f) = (2, 3)$ . Radius: $\sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$ .

Distance from centre to $P$ : $\sqrt{(5-2)^2 + (6-3)^2} = \sqrt{9 + 9} = 3\sqrt{2} \approx 4.24$ .

Since $3\sqrt{2} < 5$ (radius), $P$ is indeed inside the circle. Power is negative for interior points.

\text{Power} = \boxed{-7}

The length of the tangent from an external point is $\sqrt{S_1}$ . Since $S_1 = -7 < 0$ , no real tangent can be drawn from $P$ to the circle. This makes sense — you can’t draw a tangent to a circle from a point inside it.

If $P$ were outside (say $P(7, 8)$ ): $S_1 = 49 + 64 - 28 - 48 - 12 = 25$ , and tangent length $= \sqrt{25} = 5$ .

Why This Works

The power of a point $P$ with respect to a circle measures the “signed distance relationship” between $P$ and the circle:

Power > 0: $P$ is outside the circle → tangent length $= \sqrt{\text{power}}$
Power = 0: $P$ is on the circle
Power < 0: $P$ is inside the circle → no real tangent exists

For an external point, if a secant through $P$ intersects the circle at $A$ and $B$ , then $PA \cdot PB =$ power. If a tangent from $P$ touches at $T$ , then $PT^2 =$ power. This gives the tangent-secant relationship: $PT^2 = PA \cdot PB$ .

The formula $S_1$ is just the algebraic shortcut for computing $d^2 - r^2$ (where $d$ is the distance from $P$ to the centre).

Alternative Method — Using the distance formula directly

Power $= d^2 - r^2 = (3\sqrt{2})^2 - 5^2 = 18 - 25 = -7$ .

The power of a point is one of the most versatile tools in circle geometry for JEE. It connects tangent length, secant products, and the radical axis into one unified concept. When a problem involves tangents or secants from an external point, compute the power first — it often simplifies the entire problem.

Common Mistake

Students compute $S_1$ , get a negative value, and still try to take its square root for the “tangent length.” $\sqrt{-7}$ is not real — this should immediately signal that the point is inside the circle. Always check the sign of $S_1$ before computing tangent length. Positive → external, zero → on the circle, negative → internal.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the distance formula directly

Common Mistake

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