This isn't about your question but try using the math tab to make things less confusing. Looks like a fancy E.
See?
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?
Thanks in advance
1. If radius of A must be tangent to B that means
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
Then the tangent in T to B has the equation:
and the perpendicular bisector of TF has the equation
Calculate the x-coordinate of the point of intersection between these two lines, that means solve for x:
which yields . 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.
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:
Now subtract columnwise:
This result is valid for all circles A which satisfy the given conditions.