How do you write the standard from of the equation of the circle given center is at (-1,2) and the point (3,4) lies on the circle?

1 Answer
Mar 9, 2018

Answer:

#(x+1)^2+(y-2)^2=20#

Explanation:

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

#•color(white)(x)(x-a)^2+(y-b)^2=r^2#

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

#"here "(a,b)=(-1,2)#

#"the radius is the distance from the centre to a point"#
#"on the circumference"#

#"to calculate r use the "color(blue)"distance formula"#

#•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#"let "(x_1,y_1)=(-1,2)" and "(x_2,y_2)=(3,4)#

#rArrr=sqrt((3+1)^2+(4-2)^2)=sqrt(16+4)=sqrt20#

#rArr(x-(-1))^2+(y-2)^2=(sqrt20)^2#

#rArr(x+1)^2+(y-2)^2=20larrcolor(red)"equation of circle"#