# coordinate geometry

#### prasum

find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0

#### earboth

MHF Hall of Honor
find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0
1. The circle $$\displaystyle c_1: x^2+y^2=4^2$$ has the center (0, 0) and the radius 4.
The circle $$\displaystyle c_2:x^2+y^2-12x+32=0~\implies~(x-6)^2+y^2=2^2$$ has the center (6, 0) and the radius 2.

2. Use similar right triangles to determine the missing lengthes. I used Euklid's theorem to calculate the coordinates of the tangent points: $$\displaystyle T_1\left(\frac43\ ,\ \frac83 \sqrt{2}\right)$$

and $$\displaystyle T_2\left(\frac{20}3\ ,\ \frac43 \sqrt{2}\right)$$

3. The center of the resulting circle $$\displaystyle c_r$$ must be on the x-axis and on the perpendicular bisector of $$\displaystyle \overline{T_1T_2}$$
The slope of this perpendicular bisector equals the slope of $$\displaystyle \overline{OT_1} = \dfrac{\frac83 \sqrt{2}}{\frac43}=2\sqrt{2}$$

The midpoint of $$\displaystyle \overline{T_1T_2}$$ has the coordinates $$\displaystyle M\left(\dfrac{\frac43 + \frac{20}3}2\ , \ \dfrac{\frac83 \sqrt{2} + \frac{4}3 \sqrt{2}}2 \right)$$. That means $$\displaystyle M(4, 2\sqrt{2})$$

4. The perpendicular bisector through M has the equation:

$$\displaystyle y-2 \sqrt{2} = 2\sqrt{2}(x-4)$$

This line crosses the x-axis at:

$$\displaystyle -2 \sqrt{2} = 2\sqrt{2}(x-4)~\implies~-1=x-4~\implies~x=3$$

5. The center of the resulting circle is C(3, 0). Now calculate the distance $$\displaystyle |\overline{CT_1}|$$ to get the radius of $$\displaystyle c_r$$. I've got $$\displaystyle r=\sqrt{17}$$. Therefore the equation of $$\displaystyle c_r$$ is:

$$\displaystyle c_r: (x-3)^2+y^2=17$$

#### Attachments

• 22.5 KB Views: 51
• prasum

#### bjhopper

Thanks earboth a nice solution

bjh

thanks man

#### prasum

how did yu use euklids geometry here and what is it basically

#### earboth

MHF Hall of Honor
how did yu use euklids geometry here and what is it basically
1. I learned that in English Euclid's theorem is related to number theory. In Germany a group of theorems is summed under the name Euclid's theorems where this theorem belongs to:

In a right triangle the height divides the hypotenuse into two segments, p and q. Then

$$\displaystyle h^2 = p \cdot q$$

And that is what I used:

2. Take $$\displaystyle c_1$$ and the corresponding right triangle with

$$\displaystyle c = 12, b = 4, a = 8 \sqrt{8}$$

According to the the attached sketch the tangent point $$\displaystyle T_1$$ has the coordinates $$\displaystyle T_1(x, h)$$

3. From the area of the right tringle you know:

$$\displaystyle A=\frac12 \cdot c \cdot h = \frac12 \cdot a \cdot b ~\implies~h = \dfrac{a \cdot b}{c}$$

4. With y = 12 - x you get:

$$\displaystyle x(12-x)=h^2$$

Since you know already the value of h you can solve the equation for x.

5. Alternatively you can plug in the value of h² instead of y² into the equation of the circle to get the x-coordinate of the tangent point:

$$\displaystyle x^2+\dfrac{16 \cdot 128}{144} = 16~\implies~ |x|=\dfrac34$$

#### Attachments

• 5.6 KB Views: 23
Similar Math Discussions Math Forum Date
Geometry
Geometry
Geometry
Geometry