Find the equation of the circle with center at (3,0) and tangent to the line x-y-2=0?

1 Answer
Feb 17, 2018

#(x-3)^2+y^2=1/2#

Explanation:

#"the equation of a circle in standard form is "#

#color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))#

#"where "(a,b)" are the coordinates of the centre and r is"#
#"the radius"#

#"since "x-y-2=0" is a tangent to the circle"#

#"the distance from the centre to the point of contact"#
#"with the tangent is the radius"#

#"the distance from a line "Ax+By+C=0#
#"to a point "(m,n)" is calculated using"#

#d=(|Am+Bn+C|)/(sqrt(A^2+B^2))#

#"here "A=1, B=-1, C=-2" and "(m,n)=(3,0)#

#rArrr=(|3+0-2|)/(sqrt(1^2+(-1)^2))=1/sqrt2#

#rArr(x-3)^2+(y-0)^2=(1/sqrt2)^2#

#rArr(x-3)^2+y^2=1/2larrcolor(red)"equation of circle"#
graph{((x-3)^2+y^2-1/2)(y-x+2)=0 [-10, 10, -5, 5]}