Question

The tangent lines to the circle \[{x^2} + {y^2} - 6x + 4y = 12\] which are parallel to the line \[\] \[4x + 3y + 5 = 0\;\] are given by
A.\[3y + 4x - 31 = 0,{\text{ }}\;3y + 4x + 19 = 0\]
B.\[3y + 4x + 25 = 0,{\text{ }}\;3y + 4x - 25 = 0\]
C.\[3y + 4x - 13 = 0,{\text{ }}\;3y + 4x + 17 = 0\]
D.None

Answer 1

Hint: To solve this type of problem we will use the formula of tangent to the circle. In place of slope we will put the value of slope of parallel line given in this question. Because parallel lines always have the same slope.

Complete step-by-step answer:
Given the equation of circle
\[{x^2} + {y^2} - 6x + 4y = 12\]
So the centre of circle \[ = \left( { - g, - f} \right) = \left( {3, - 2} \right)\]
and radius \[r = \sqrt {{g^2} + {f^2} - c} = \sqrt {9 + 4 + 12} = 5\]
We know the equation of tangent on a circle is
\[y = mx \pm\times r \sqrt {1 + {m^2}} \] …….(i)

Here the value of m will be equal to the slope of the given parallel line
Therefore we need to find the slope of the given parallel line
\[
\Rightarrow {4x + 3y + 5 = 0} \\
\Rightarrow {3y = - 5 - 4x} \\
\Rightarrow {y = \dfrac{{ - 5}}{3} - \dfrac{{4x}}{3}}
\]
So here slope \[ = \dfrac{{ - 4}}{3}\]
So from (i) tangent equation
\[
\Rightarrow {y = \dfrac{{ - 4x}}{3} \pm 5\sqrt {1 + \dfrac{{16}}{9}} } \\
\Rightarrow {y = \dfrac{{ - 4x}}{3} \pm 5 \times \dfrac{5}{3}}
\]
\[3y = - 4x \pm 25 \Rightarrow \;\] so here the equation of tangents are \[3y + 4x + 25 = 0,{\text{ }}\;3y + 4x - 25 = 0\]
So, the correct answer is “\[3y + 4x + 25 = 0,{\text{ }}\;3y + 4x - 25 = 0\] ”.

Note: In this type of problem there will always be a pair of tangents on the circle and these tangents are parallel to each other use this concept to proceed with the solution and get the required.