Question

Consider the circle $x^2+y^2-6x+4y=12$. Then one of the equations of a tangent to this circle that is parallel to the line $4x+3y+5=0$ is :

• A. $4x+3y+10=0$
• B. $4x+3y-9=0$
• C. $4x+3y+9=0$
• D. $4x+3y-31=0$

Answer 1

The correct option is D $4x+3y-31=0$

$x^2+y^2-6x+4y-12=0$

=> $(x-3{)}^2+(y+2{)}^2=5^2$

This is a circle whose centre is $(3,-2)$ and radius $5$

Any line parellel to $4x+3y+5=0$ can be written in form $4x+3y+c=0$ (as the slope of both lines is same)

Therefore, let the tangent to the circle be $4x+3y+c=0$

Now, we need to find out value of c for which this line is tangent to the given circle. We know that lenght of tangent from the centre of the circle is equal to it's radius.

=> perpendicular distance of $4x+3y+c=0$ from $(3,-2)$ should be equal to 5

Applying the formulae for perpendicular distance of a point from a line, we get

$\frac{|4\left(3\right)+3(-2)+c|}{\sqrt{4^2+3^2}}=5$

=> $|6+c|=25$

=>$6+c=25;6+c=-25$

=> $c=19,-31$

Therefore, equation of tangents are $4x+3y+19=0;4x+3y-31=0$