Question

Find the standard form of the equation of

the circle.
Center (4, -2) and tangent to the line x = 1 Please explain

Answer 1

Answer: (x-4)^2+(y+2)^2=18

Explanation:

First fine line perpendicular to x=1(the opposite reciprocal of 1)

x2=-1*1/x

x2=-1/x

x2=-1/1

x2=-1 ==> perpendicular line

y=-x+b

-2=-4+b

b=2

y=-x+2

x2=x ==> Set x and x2 to equal each other to find the value of x.

-x+2=x

2x=2

x=1

y=-x+2

y=-1+2

y=1

Circle equation: (h-x)^2+(k-y)^2=r^2.

(-2-y)^2+(4-x)^2=r^2

(-2-1)^2+(4-1)^2=r^2

(-3)^2+3^2=r^2

9+9=r^2

r^2=18

r=18^1/2

(x-4)^2+(y-(-2))^2=r^2

(x-4)^2+(y+2)^2=18