Question

The center of a circle and a point on the circle are given. Write the equation of the circle in standard form.

Answer 1

$(x-2)^2+(y+2)^2=34$

Step-by-step explanation:

Write the equation of the circle in standard form, where center: (2,-2), point on the circle: (5,3)

The equation of circle in standard form is:

$r^2=(x-h)^2+(y-k)^2$

Where (h,k) is center and r is the radius.

We are given center (2,-2) and we need to find radius r

We know a point on circle (5,3) and center (2,-2)

Finding distance between these points:

$r = d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \\r= d=\sqrt{(5-2)^2+(3-(-2))^2}\\r= d=\sqrt{(3)^2+(5)^2}\\r= d=\sqrt{9+25}\\r= d=\sqrt{34}$

So, r = √34

then r^2 = 34

So, equation of circle is:

$(x-h)^2+(y-k)^2=r^2\\(x-2)^2+(y-(-2))^2=34\\(x-2)^2+(y+2)^2=34$

Keywords: Equation of circle

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