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​ _✨⭐

❌❌ _ UNKNOWN PEOPLE DO NOT GIVE ME THANKS PLEASE _ ❌❌​

Answer 1

ok dear but I am not unknown na 4 u

The tgt at (— 1,2) yo the circle

x^2+y^2-4x- 8y+7= 0 is

- x+2y -2( x — 1) — 4( y+2)+7 = 0

that is — 3x — 2y +1 = 0 that is 3x+2y- 1 = 0

The center and radius of the other circle

are C( — 2,— 3,) and 13^1/2

The length of perpendicular fro the center C to the line 3x+2y- 1 = 0 is | — 13|/(13^(1/2)

= 13^1/2 = radius

So the tangent line 3x+2y- 1 = 0

touches the circle x^2+y^2+4x +6y = 0

Te foot of perpendicular from C( -2,-3) on the line 3x+2y- 1 = 0 is the point of contact and is given by

(x +2)/3 = ( y+3)/2 = — ( -6-6-1)/13= 1

So x = -2 +3 = 1 and y = -3+2= -1

So the point of contact is ( 1 ,—1)

Answer 2

Hope it helps uh...

keep smiling

Thank uh...!!

plz don't report my answer, I tried my best to give correct answer!!