Question

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)?

Answer 1

`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`