# Finding a point of intersection between circles

• Apr 13th 2010, 01:29 PM
precalc1209
Finding a point of intersection between circles
My teacher has us going through these A level problems, but I am just so lost with this one. She said there may be a pop quiz tomorrow with similar questions, so i'd really like to get the concepts down flat. Would anyone be willing to explain how to solve the following word problem?

A) The circles (x^2)+(y^2)=20 and ((x-6)^2)+((y-3)^2)=5 are tangent. Find the coordinates of the point of trangency and then find an equation of the common internal tangent shown.

B) If you subtract the equations of the two circles from each other, you get a linear equation. Find this quation. Compare it with the quation found in part A.

Any help is appreciated. Thank you!

-Rob
• Apr 13th 2010, 01:44 PM
Quote:

Originally Posted by precalc1209
My teacher has us going through these A level problems, but I am just so lost with this one. She said there may be a pop quiz tomorrow with similar questions, so i'd really like to get the concepts down flat. Would anyone be willing to explain how to solve the following word problem?

A) The circles (x^2)+(y^2)=20 and ((x-6)^2)+((y-3)^2)=5 are tangent. Find the coordinates of the point of trangency and then find an equation of the common internal tangent shown.

B) If you subtract the equations of the two circles from each other, you get a linear equation. Find this quation. Compare it with the quation found in part A.

Any help is appreciated. Thank you!

-Rob

Hi precalc1209,

the centre of $\displaystyle x^2+y^2=\left(\sqrt{20}\right)^2$ is (0,0)

radius is $\displaystyle \sqrt{20}=\sqrt{5(4)}=\sqrt{5}\sqrt{4}=2\sqrt{5}$

the centre of $\displaystyle (x-6)^2+(y-3)^2=\left(\sqrt{5}\right)^2$ is (6,3)

radius is $\displaystyle \sqrt{5}$

Distance from (0,0) to (6,3) is $\displaystyle \sqrt{6^2+3^2}=\sqrt{45}=\sqrt{9(5)}=\sqrt{9}\sqrt {5}=3\sqrt{5}$

hence the circles touch 2 thirds of the way from (0,0) to (6,3) which is (4,2)

the equation of the line from (0,0) to (6,3) is $\displaystyle y-3=0.5(x-6)$

the tangent line at the point of contact is perpendicular to this line
(slopes multiply to give -1) and contains the point (4,2)

You can then write the equation of this line.

now subtract the circle equations having multiplied out the factors in one of them.