Find the equation of circle with centre (1,-2),touching the line 4y-3x=5 using x²+y²-2xh-2ky+c as the general formula (answer=25x-50x+100y-131)?

1 Answer
Sep 17, 2015

25x^2+25y^2-50x+100y-131=0

Explanation:

Firstly we can find the perpendicular distance of the straight line 4y-3x=5 from center (h,k)=(1,-2) by using the equation;

distance=|ah+bk+c|/sqrt(a^2+b^2)

The value of a, b, and c can be obtained from 4y-3x=5;

4y-3x=5

Rearrange, 3x-4y+5=0

a=3 , b=-4 and c=5

And we will get;

distance=|3(1)+(-4)(-2)+5|/sqrt((3)^2+(-4)^2)

distance=16/5

You will then figure out that the distance=16/5 is the radius, r of the circle. By substituting r=16/5 and (h,k)=(1,-2) , you can use the equation;

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

(x-1)^2+(y+2)^2=(16/5)^2

Expand and you will get;

x^2-2x+1+y^2+4y+4=256/25

All values multiply by 25;

25x^2-50x+25+25y^2+100y+100=256

25x^2+25y^2-50x+100y-131=0