Question

The length of the tangent from the point \((5,1)\) to the circle \(x^2+y^2+6x-4y-3=0\) is

• A. \(81\)
• B. \(7\)
• C. \(29\)
• D. \(21\)

Hint:

For a circle \(S=0\), the tangent length from a point \((x_1,y_1)\) is \(\sqrt{S_1}\), where \(S_1\) is obtained by substituting the point in the circle equation.

Answer 1

The Correct Option is B

We are given the circle \[ x^2+y^2+6x-4y-3=0. \] The length of tangent from an external point \((x_1,y_1)\) to the circle \[ S=0 \] is given by \[ \sqrt{S_1}, \] where \(S_1\) is obtained by substituting \((x_1,y_1)\) in the equation of the circle. Here, the external point is \[ (5,1). \] So, \[ x_1=5,\qquad y_1=1. \] Now substitute \(x=5\) and \(y=1\) in the circle: \[ S_1=(5)^2+(1)^2+6(5)-4(1)-3. \] Now simplify step by step: \[ S_1=25+1+30-4-3. \] \[ S_1=26+30-7. \] \[ S_1=56-7. \] \[ S_1=49. \] Therefore, the length of the tangent is \[ \sqrt{S_1}=\sqrt{49}. \] \[ =7. \] Hence, the length of the tangent is \[ 7. \]