Question

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

Answer 1

The given circles be

S: x² + y² − 4x − 8y + 7 = 0

-

S': x² + y² + 4x+6y=0

The equation of tangent at (-1,2) is T = 0

i.e.,

T: (−1)x + (2)y − 2(−1+x) − 4(2 + y) + 7

T: − x +2y+2 − 2x −8 − 4y + 7 = 0

T: 3x - 2y + 1 = 0

T: 3x +2y-1=0

Now, centre of S' be C' and radius be r'

SO,

C'= (-2,-3) r² = √√4+9= √13

Perpendicular distance from (-2,-3) to T=0 will be the radius,

So,

d |-6-6-1| √9+4 13 √13 = = √13

Since, d=r'

hence 3x+2y-1=0 touches S'=0