Question

find the equations of the tangents to the circle x^2+y^2+2x-2y-3=0 which are perpendicularto 3x-y+4=0​

Answer 1

ANSWER

(i)

Let the Equation of tangent parallel to 3x−4y−1=0 be 3x−4y+k=0

Now, Radius = Distance of center from the line

$\sqrt{ {1}^{2} \div {2}^{2} - (4)} = | \frac{3(1)−4(2)+k</p><p>}{ \sqrt{9 + 16} } |$

⇒3∗5=3−8+k⇒k=20

Hence, Equation of tangent parallel to 3x−4y−1=0

is 3x−4y+20=0

(ii)

Let the Equation of tangent perpendicular to 3x−4y−1=0 be 4x+3y+p=0

Now, Radius = Distance of center from the line

$\sqrt{ {1}^{2} \div {2}^{2} - (4)} = | \frac{4(1)+3(2)+p}{ \sqrt{16 \div 9} } |$

⇒3∗5=4+6+p⇒p=5

Hence, Equation of tangent perpendicular to 3x−4y−1=0

is 4x+3y+5=0