Question

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

Answer 1

`(x+6)^2+(y-2)^2=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"`

`"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"`

Answer 2

`(x+6)^2+(y-2)^2=2`

Explanation:

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

Given the center `(-6,2)` and the point `(-5,3)`
The distance between the points parallel to the x-axis is `1` (the absolute value of the difference between the x-coordinate values), and
the distance between the points parallel to the y-axis is also `1` (the absolute value of the difference between the y-coordinate values).

By the Pythagorean Theorem this distance (the radius, remember) is `r=sqrt(1^2+1^2)=sqrt(2)`

The general equation for a circle with center `(a,b)` and radius `r` is
`color(white)("XXX")(x-a)^2+(y-b)^2=r^2`

So with a center of `(-6,2)` and a radius of `sqrt(2)`,
we have
`color(white)("XXX")(x+6)^2+(y-2)^2=2`