1. ## Circles

Problem: A circle A passing through the point (4, 0) is orthogonal to a circle B with equation x^2+y^2=4. Show that the locus of the centre of the circle A is a straight line and find its equation.

Well, I first established the centre and radius of circle B as (0, 0) and 2 respectively. Then, I let the centre of circle A be (a,b). Then, I reasoned that since circles are orhtogonal, the square of the distance between the centre should be equal to the sum of the squares of the two radii, so that: a^2+b^2 = 4 + (0-a)^2 + (4-b)^2. This gave me the value of a but then I got stuck. Was I on the right track? Any suggestions please?

2. This isn't about your question but try using the math tab to make things less confusing. Looks like a fancy E. $x^2+y^2=4$

See?

3. Originally Posted by yobacul
Problem: A circle A passing through the point F(4, 0) is orthogonal to a circle B with equation x^2+y^2=4. Show that the locus of the centre of the circle A is a straight line and find its equation.

...
1. If $A \perp B~\implies~$ radius of A must be tangent to B that means $A \perp B~\implies~r_A \perp r_B$

2. If C(a, b) as you suggested then C is the point of intersection between the tangent in T and the perpendicular bisector of TF.

3. I've attached a drawing of the situation. The locus of all C is drawn in red.

EDIT: I used a slightly different approach

Let $T(t, \sqrt{4-t^2})$

Then the tangent in T to B has the equation:

$y = \dfrac{4-tx}{\sqrt{4-t^2}}$ and the perpendicular bisector of TF has the equation $y = -\dfrac{x(4-t)+6}{\sqrt{4-t^2}}$

Calculate the x-coordinate of the point of intersection between these two lines, that means solve for x:

$\dfrac{4-tx}{\sqrt{4-t^2}} = -\dfrac{x(4-t)+6}{\sqrt{4-t^2}}$ which yields $x = \dfrac52$ . As you can see there isn't a parameter t anymore. So x = 2.5 is the equation of the locus of the centers of all possible circles A.

4. ## Still confused ...

Many thanks for answering the question. I tried to go through all the steps but I am still confused. I understood why T takes those coordinates but then I got lost.

5. Originally Posted by yobacul
Many thanks for answering the question. I tried to go through all the steps but I am still confused. I understood why T takes those coordinates but then I got lost.
I've found a much simpler way to solve this question:

1. See attachment. Actually you are dealing with 3 right triangles, labeled I, II and III.

2. Let r denote the radius of circle B and R denote the radius of circle A. C(a, b) is the center of circle A.

3. Then you get a system of equations:

$\left|\begin{array}{rcl}r^2+R^2&=&a^2+b^2\\ R^2&=&(4-a)^2+b^2\end{array} ~\implies~\right.$ $\left|\begin{array}{rcl}4+R^2&=&a^2+b^2\\ R^2&=&16-8a+a^2+b^2\end{array} \right.$

Now subtract columnwise:

$4 = -16 +8a~\implies~\boxed{a=\dfrac52}$

This result is valid for all circles A which satisfy the given conditions.