Find the length of the tangent from the point (5, 1) to the circle x2 + y2 + 6x - 4y - 3 = 0 ?

CONCEPT:

The length of the tangent from an external point P (x1, y1) to the circle represented by the equation: x2 + y2 + 2gx + 2fy + c = 0 is given by: \(\sqrt {x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c} \)

CALCULATION:

Given: Equation of circle is x2 + y2 + 6x - 4y - 3 = 0 and point (5, 1)

If we substitute x = 5 and y = 1 in the expression x2 + y2 + 6x - 4y - 3, we get

⇒ 52 + 12 + 6 × 5 - 4 × 1 - 3 = 49 > 0---------(1)

So, the point (5, 1) is an external point.

As we know that, the length of the tangent from an external point P (x1, y1) to the circle represented by the equation: x2 + y2 + 2gx + 2fy + c = 0 is given by: \(\sqrt {x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c} \)

Here, x1 = 5 and y1 = 1 and from equation (1) we know that, x12 + y12 + 6x1 - 4y1 - 3 = 49

So, the length of the tangent is √49 = ±7 units

∵ The length cannot be negative. So, the length of tangent is 7 units

Hence, option C is the correct answer.