Question

Write an equation for a circle with its center at (-3, 4) and passes through the point (-6, 7). Leave your answer in standard form.

Answer 1

To solve this question, first, we must know the standard form for the equation of a circle.

Which is:

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

To understand this question better, let us have a pictorial view of it:

We will need to find the radius of the circle to be able to get what the equation of the circle is:

To find the radius, we will solve for the distance between points (-3.4) and (-6,7)

The formula for finding the distance between two points is:

`\begin{gathered} =\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \end{gathered}`

Substituting the given coordinate above into the formula will give us:

`\sqrt[]{(-6-(-3))^2+(7-4)^2}`
`\begin{gathered} \sqrt[]{(-6+3)^2+(3)^2} \\ \sqrt[]{(-3)^2+9} \\ \sqrt[]{9+9} \\ =\sqrt[]{18} \\ r=\sqrt[]{18} \end{gathered}`

To get the equation of the circle:

`\begin{gathered} \text{Center coordinates: (-3,4)} \\ h=-3 \\ k=4 \\ r=\sqrt[]{18} \\ \text{Substituting the values above into the standard equation of a cirlce} \\ we\text{ will have:} \\ (x-(-3))^2+(y-4)^2=(\sqrt[]{18})^2 \\ (x+3)^2+(y-4)^2=18 \\ \text{The equation above is the equation of the circle in standard form.} \end{gathered}`