# What is the circle of the equation if the center is (-6,2) passes through the point (-5,3)? Thank you for the answer.

##### 2 Answers

#### 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"#

#"here "(a,b)=(-6,2)#

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

#"the circle"#

#"calculate r using 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)=(-6,2)" and "(x_2,y_2)=(-5,3)#

#r=sqrt((-5+6)^2+(3-2)^2)=sqrt(1+1)=sqrt2#

#"substitute these values into the equation"#

#(x-(-6))^2+(y-2)^2=(sqrt2)^2#

#(x+6)^2+(y-2)^2=2larrcolor(red)"equation of circle"#

#### Explanation:

The radius is the distance from the center to any point on the circumference.

Given the center

The distance between the points parallel to the x-axis is

the distance between the points parallel to the y-axis is also

By the Pythagorean Theorem this distance (the radius, remember) is

The general equation for a circle with center

So with a center of

we have