# Circles

• Sep 16th 2010, 05:48 AM
Punch
Circles
Given the equation of a circle is $x^2+(y-6)^2=8$.

And the equation of a second circle is $x^2+y^2=6y+d$, where $d$ is an integer, is the reflection of the circle $x^2+(y-6)^2=8$ about the line $y=k$, find the values of $k$ and of $d$
• Sep 16th 2010, 05:59 AM
Unknown008
Move the 6y on the same side as y^2 and complete the square for y, to get it in the same form as your first given equation.

Since the second one is a reflection of the first one, their radius is the same, you can find d.

Then, find the midpoint of the centres of the two circles. This will give you the value of k.

Post what you get! (Happy)
• Sep 16th 2010, 06:12 AM
MathoMan
Quote:

Originally Posted by Punch
Given the equation of a circle is $x^2+(y-6)^2=8$.

And the equation of a second circle is $x^2+y^2=6y+d$, where $d$ is an integer, is the reflection of the circle $x^2+(y-6)^2=8$ about the line $y=k$, find the values of $k$ and of $d$

$x^2+y^2=6y+d\Rightarrow x^2+(y-3)^2=d+9$.

Since transformation is reflection, then both circles must have the same radius, hence $8=d+9\Rightarrow d=-1.$

Both circles have its center on the y-axis $C_1(0,6)$ and $C_2(0,3)$ so $k=\frac{6+3}{2}=\frac{9}{2}.$