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