find the equation of the circle : tangent to 3x - 4y + 3 = 0 at ( -1,0 ) ; radius 7?

Question

2 equations are possible......

Answer 1

Let the centre of the circle be (a,b).
Then the length of perpendicular from (a,b) to the line
= |(3a - 4b + 3)/{sqrt(3^2 + 4^2)}|
= + / - (3a - 4b + 3)/5

Taking + sign

This distance is the radius of the circle.
So, (3a - 4b + 3)/5 = 7
or 3a - 4b = 32
or b = (3a - 32)/4 _______ (1)

Again, distance between (a,b) and (-1,0) is 7 units
So, (a + 1)^2 + b^2 = 7^2 = 49 _________ (2)

Substituting 'b' from (1) in (2)
(a + 1)^2 + (3a - 32)^2/16 = 49
16a^2 + 32a + 16 + 9a^2 - 192a + 32^2 = 49*16
25a^2 - 160a + 256 = 0
(5a - 16)^2 = 0
a = 16/5
b = -28/5

Equation of the circle is: (x - 16/5)^2 + (y + 28/5)^2 = 7^2

Taking '-' sign and following the same procedure the other circle is:
(x + 26/5)^2 + (y - 28/5)^2 = 7^2

Answer 2

The center must be on the lines
(3x-4y+3)/(+-5) = 7
So there are 2 possibilities
3x-4y =32 and3x-4y=-38
Take the perpendicular to 3x-4y+3=0 through (-1,0)
y= -4/3(x+1) 3y+4x =- 4
I´ll solve one case
3x-4y=32
4x+3y=-4
9x-12y=96
16x+12y=-16
sum 25x= 80 and x= 16/5 y= -28/5

(x-16/5)^2+(y+28/5)^2 =49
you can do the other case

Answer 3

I asked the same question 4 times, and didn't get an answer

Answer 4

To be honest, I don't think so

Answer 5

Do your own homework.....