Question

Find the equation of the tangent to the circle x² + y²-6x-3y-2=0 at (2,-2)​

Answer 1

Answer:

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

4

2

+3

2

∣4(3)+3(−2)+c∣

=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