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The equations to the tangents to the circle x^2+y^2-6 x+4 y=12 which are parallel to the straight line

4 x+3 y+5=0, are

Option: 1

3 x-4 y-19=0, \quad 3 x-4 y+31=0


Option: 2

4 x+3 y-19=0, \quad 4 x+3 y+31=0


Option: 3

4 x+3 y+19=0, \quad 4 x+3 y-31=0


Option: 4

3 x-4 y+19=0, \quad 3 x-4 y+31=0


Answers (1)

best_answer

Let equation of tangent be 4 x+3 y+k=0, then \sqrt{9+4+12}=\left|\frac{4(3)+3(-2)+k}{\sqrt{16+9}}\right| \Rightarrow

6+k= \pm 25 \Rightarrow k=19 \, \, and -31

Hence the equations of tangents are 4 x+3 y+19=0 \text { and } 4 x+3 y-31=0

Posted by

HARSH KANKARIA

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