Circles

**Punch** · September 16th 2010, 05:48 AM

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 $d$

**Unknown008** · September 16th 2010, 05:59 AM

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!

**MathoMan** · September 16th 2010, 06:12 AM

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 $k=\frac{6+3}{2}=\frac{9}{2}.$

Thread: Circles

LinkBack

Thread Tools

Display

Circles

Similar Math Help Forum Discussions

Prove that every rigid motion transforms circles into circles

How much of the area is preserved when you inscribe circles in circles

4 circles

going in circles

Circles - Please Help

Search Tags