Tangent from external point to circle — find length and equation

Question

Find the length and equations of the tangents drawn from the point $P(5, 4)$ to the circle $x^2 + y^2 - 4x - 6y + 8 = 0$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

$x^2 + y^2 - 4x - 6y + 8 = 0$

Complete the square: $(x-2)^2 + (y-3)^2 = 4 + 9 - 8 = 5$

Centre $C = (2, 3)$ , radius $r = \sqrt{5}$ .

For a point $P(x_1, y_1)$ outside the circle $S = 0$ :

\text{Length} = \sqrt{S_1}

where $S_1 = x_1^2 + y_1^2 - 4x_1 - 6y_1 + 8$ .

S_1 = 25 + 16 - 20 - 24 + 8 = 5

\text{Length of tangent} = \sqrt{5}

Let the tangent from $P(5, 4)$ have slope $m$ . Its equation: $y - 4 = m(x - 5)$ , i.e., $mx - y + (4 - 5m) = 0$ .

For this to be tangent to the circle, the perpendicular distance from centre $(2, 3)$ must equal $r = \sqrt{5}$ :

\frac{|2m - 3 + 4 - 5m|}{\sqrt{m^2 + 1}} = \sqrt{5}

\frac{|1 - 3m|}{\sqrt{m^2 + 1}} = \sqrt{5}

Squaring: $(1 - 3m)^2 = 5(m^2 + 1)$

$1 - 6m + 9m^2 = 5m^2 + 5$

$4m^2 - 6m - 4 = 0$

$2m^2 - 3m - 2 = 0$

$(2m + 1)(m - 2) = 0$

$m = 2$ or $m = -\frac{1}{2}$

Tangent 1: $y - 4 = 2(x - 5)$ $\Rightarrow$ $\mathbf{2x - y - 6 = 0}$

Tangent 2: $y - 4 = -\frac{1}{2}(x - 5)$ $\Rightarrow$ $\mathbf{x + 2y - 13 = 0}$

Why This Works

From any external point, exactly two tangents can be drawn to a circle. The length formula $\sqrt{S_1}$ comes from the Pythagorean theorem: in the right triangle formed by the centre, the external point, and the point of tangency, the tangent length is the third side.

The tangent condition (distance from centre = radius) converts the geometry into algebra. Two values of $m$ correspond to the two tangent lines.

Notice the two tangent slopes are $2$ and $-1/2$ — their product is $-1$ . This means the two tangents are perpendicular. This happens when the external point lies on the director circle ( $x^2 + y^2 = 2r^2$ for a circle centred at origin).

Alternative Method

Use the equation of tangent in terms of slope for the standard circle: $y = mx + c$ is tangent to $(x-h)^2 + (y-k)^2 = r^2$ when $c = k - mh \pm r\sqrt{1+m^2}$ . This gives the same result but is more formula-heavy.

A quick check: the tangent length squared ( $S_1 = 5$ ) equals $r^2 = 5$ . When tangent length = radius, the two tangents from $P$ are perpendicular to each other. This is a useful shortcut for MCQs that ask about the angle between tangents.

Common Mistake

When using the slope form, students often forget to check if one tangent is vertical ( $m \to \infty$ ). In this problem, the quadratic in $m$ gives two finite slopes, so no issue. But if the quadratic gives only one finite root, check whether a vertical tangent ( $x = 5$ ) also works. Missing the vertical tangent means missing one of the two answers.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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