Question

Find an equation (s) of circle (s) tangent to line 3x-4y-4=0 at point (0,-1) and containing the point (-1,-8)​

Answer 1

Answer:

$x^{2}$+$y^{2}$-6x+10y+9=0

Step-by-step explanation:

First, get the slope of 3x-4y-4=0:

3x-4y-4=0

4y=3x-4

y=3x/4-1

m=3/4

Then, get the perpendicular line of 3x-4y-4=0:

slope of the perpendicular line is -4/3.

using (0,-1) we will use point - slope formula:

Then,

y-$y_{1}$=m(x-$x_{1}$)

y+1 = -4/3 (x-0)

4x+3y+3=0   eq (1)

Then, connect the points (0,-1) and (-1,-8)

We will determine the midpoint of (0,-1) and (-1,-8)

$\frac{-1+0}{2}$= -1/2=x

$\frac{-8-(-1)}{2}$= -9/2=y

We will determine the slope of these points.

m=$\frac{-1-(-8)}{0-(-1)}$=7

Then we will get the perpendicular line slope

Therefore,

m= -1/7

Also, we will get the perpendicular line using point slope formula using

m= -1/7 and the midpoint (-1/2,-9/2)

y-$y_{1}$=m(x-$x_{1}$)

y+9/2= -1/7(x+1/2)

x+7y= - 32      eq(2)

Using eq (1) and (2) to get the center of the circle.

x+7y= -32 eq (2)

4x+3y= -3 eq (1)

By Elimination method:

We get:

x= 3

y= -5

Then,

C(h,k) = C (3,-5);

Then,

we will get the radius of the circle using distance formula

For this we will use the C (3,-5) and (-1,-8).

r=$\sqrt{(3+1)^{2}+(-5+8)^{2} }}$

r=5

Substitute,

(x-h)^2 + (y-k)^2=r^2

(x-3)^2  + (y + 5)^2=5^2

$x^{2} -6x +9 + y^{2} +10y +25=25$

Ans:

$x^{2} +y^{2} -6x +10y +9=0$