Answer
$\dfrac{(x-1)^2}{9}+\dfrac{(y-2)^2}{9}=1$
See graph (circle)
Work Step by Step
We are given the ellipse:
Center: $(1,2)$
Vertex: $(4,2)$
Point on the graph: $(1,5)$
Because the $y$-coordinates of the center and focus are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k$ using the center:
$(h,k)=(1,2)$
$h=1$
$k=2$
Determine $a$ using the vertex:
$(h+a,k)=(4,2)$
$(1+a,2)=(4,2)$
$1+a=4$
$a=3$
Determine $b$ using the point $(1,5)$:
$\dfrac{(1-1)^2}{a^2}+\dfrac{(5-2)^2}{b^2}=1$
$\dfrac{9}{b^2}=1$
$b^2=9$
$b=3$
The equation of the ellipse is:
$\dfrac{(x-1)^2}{3^2}+\dfrac{(y-2)^2}{3^2}=1$
$\dfrac{(x-1)^2}{9}+\dfrac{(y-2)^2}{9}=1$
As $a=b$ the equation represents a circle.
The center is:
$(h,k)=(1,2)$
Determine the vertices and co-vertices:
$(h-a,k)=(1-3,2)=(-2,2)$
$(h+a,k)=(4,2)$
$(h,k-b)=(1,2-3)=(1,-1)$
$(h,k+b)=(1,2+3)=(1,5)$
Graph the circle:
