What is the general form of the equation of a circle whose center is (-4,-3) and contains the point (-3,3)? a). x^2 + y^2 – 6x + 6y - 12 = 0 b). x^2 + y^2 + 6x - 6y - 17 = 0 c). x^2 + y^2 + 6x + 8y - 17 = 0 d). x^2 + y^2 +8x + 6y - 12 = 0

1 Answer
Mar 9, 2018

#x^2+y^2+8x+6y-12=0to(d)#

Explanation:

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

#•color(white)(x)(x-a)^2+(y-b)^2=r^2#

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

#"here "(a,b)=(-4,-3)#

#" 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)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

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

#r=sqrt((-3+4)^2+(3+3)^2)=sqrt(1+36)=sqrt37#

#(x-(-4)^2+(y-(-3)^2=(sqrt37)^2#

#rArr(x+4)^2+(y+3)^2=37larrcolor(red)"equation of circle"#

#"the equation of a circle in "color(blue)"general form"# is.

#•color(white)(x)x^2+y^2+2gx+2fy+c=0#

#"to obtain this form expand the equation of the circle"#
#"and rearrange"#

#rArr(x+4)^2+(y+3)^2=37#

#rArrx^2+8x+16+y^2+6y+9-37=0#

#rArrx^2+y^2+8x+6y-12=0larrcolor(red)"general equation"#