Question

Find the equations of the circles passing through (1, -1)and touching the lines
4x + 3y + 5 = 0 and 3x – 4y – 10 = 0

Answer 1

Answer:

hey!

y=c-3x is the tangent to a circle: x^2+y^2-4x-2y-5=0

x^2+(c-3x)^2 -4x-2(c-3x)-5 =0

x^2+9x^2-6cx+c^2 -4x-2c+6x-5 =0

10x^2+(2-6c)x+c^2-2c-5 =0

has only one solution (tangent)if:

(2-6c)^2-4(10)(c^2-2c-5) = 0

-4c^2+56c+204 =0

c^2-14c-51 =0

c = (14+20)/2 or c = (14-20)/2

Answer: c =17 or c = -3

hope it will be helpful ✌️

Answer 2

Answer:

$hello$

heya friend

here is your answer

===========================

let,

3x-4y = 10 ----------> {1}

4x+3y = 5 ------------> {2}

{1} × 3 = 9x - 12y = 30

{2} ×4 = 16x + 12y = 20

====================

= 25x = 50

x = 50/25

x = 2

therefore , x = 2

now place x in eq 1

(3×2) - 4y = 10

6 - 4y = 10

-4y = 4

y = (-1)

sho ,

x = 2

y = (-1)

i hope it helped uh !!

thanks