Question

show that the tangent at (-1,2)of the circle x^2+y^2-4x-8y+7=0 touches the circle x^2+y^2+4x+6y=0 and olso find its point of contact

Answer 1

we have to show that tangent at (-1, 2) of the circle x² + y² - 4x - 8y + 7 = 0 touches the circle x² + y² + 4x + 6y = 0 and also find its point of contact.

solution : first find equation of tangent at (-1,2) of the circle x² + y² - 4x - 8y + 7 = 0.

centre of circle = (2, 4)

so, slope of normal = (4 - 2)/(2 + 1) = 2/3

so slope of tangent = -1/slope of normal

= -1/(2/3) = -3/2

now equation, (y - 2) = -3/2(x + 1)

⇒2(y - 2) + 3(x + 1) = 0

⇒3x + 3 + 2y - 4 = 0

⇒3x + 2y - 1 = 0 ..........(1)

line 3x + 2y - 1 = 0, touches the circle x² + y² + 4x + 6y = 0 only if distance between centre to line = radius of the circle.

here centre = (-2, -3) and radius = √(2² + 3²) = √13

⇒LHS = distance between centre to line

= |3(-2) + 2(-3) - 1|/√(2² + 3²)

= |-6 - 6 - 1|/√13

= √13

RHS = radius of circle = √13

LHS = RHS

Therefore tangent at (-1,2)of the circle x² + y² - 4x - 8y + 7 = 0 touches the circle x² + y² + 4x + 6y = 0.

now find point of contact,

solving equations, 3x + 2y - 1 = 0 and x² + y² + 4x + 6y = 0

we get, x = 1 and y = -1

so, the point of contact is (1, -1)

Answer 2

Step-by-step explanation:

show that the tangent at (-1,2)of the circle x^2+y^2-4x-8y+7=0 touches the circle x^2+y^2+4x+6y=0 and olso find its point of contact