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##

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