Question

The straight line 3x – 4y + 7 = 0 is a tangent to the circle
x2 + y2 + 4x + 2y + 4 = 0 at P; find the equation of its normal at the same point.​

Answer 1

EXPLANATION.

Straight lines 3x - 4y + 7 = 0 is tangent to the

circle = x² + y² + 4x + 2y + 4 = 0.

To find the equation of its normal at

same point.

Equation = 3x - 4y + 7 =0.

Equation of circle = x² + y² + 4x + 2y + 4 = 0.

Center of circle = ( -g, -f)

Radius of circle = √(g²) + (f²) - c = 0

General equation of circle

→ x² + y² + 2gx + 2fy + c = 0

compare their equation to general equation

we get,

Center of circle = ( -2,-1).

Radius = √(-2)² + (-1)² - 4 = 0

Radius = √ 4 + 1 - 4 = 0

Radius = √1 = 1

Straight lines is perpendicular to the equation

3x - 4y + 7 = 0.

Equation of perpendicular line

-bx + ay + c = 0.

Equation = 4x + 3y + c = 0.

→ 4(-2) + 3(-1) + c = 0

→ -8 - 3 + c = 0.

→ c = 11.

Equation of its Normal.

→ 4x + 3y + 11 = 0.