To solve this question, first, we must know the standard form for the equation of a circle.
Which is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]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:
[tex]\begin{gathered} =\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \end{gathered}[/tex]Substituting the given coordinate above into the formula will give us:
[tex]\sqrt[]{(-6-(-3))^2+(7-4)^2}[/tex][tex]\begin{gathered} \sqrt[]{(-6+3)^2+(3)^2} \\ \sqrt[]{(-3)^2+9} \\ \sqrt[]{9+9} \\ =\sqrt[]{18} \\ r=\sqrt[]{18} \end{gathered}[/tex]To get the equation of the circle:
[tex]\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}[/tex]
