Problem
Let C1 be the circle of radius 1 with center at the origin. Let C2 be the circle of radius r with center at the point A(4, 1), where 1 < r < 3. Two distinct common tangents PQ and ST of C1 and C2 are drawn. The tangent PQ touches C1 at P and C2 at Q. The tangent ST touches C1 at S and C2 at T. Mid points of the line segments PQ and ST are joined to form a line which meets the x-axis at a point B. If AB = √5 , then the value of r².
Solution
Here are the two circles, and their common tangents – all four of them.
Drag the slider to change the radius of C2. Two questions:
- Why do you think the condition 1 < r < 3 is given?
- What do you notice about the midpoints of the common tangents?
I’ll discuss three different solutions, in decreasing order of complexity.
Method 1
Let’s pick the direct common tangents to proceed. Here’s a simplified figure.
Could we’ve picked any two of the four tangents? The answer’s yes – and the line joining the midpoints would’ve still passed through the same point B. More about that in the third method.
Coming back – the idea here is to find the coordinates of B in terms of r. Then, using the distance formula to equate AB to √5, we can find the value of r².
To find the coordinates of B, we’ll need the equation of CD, and for that, we’ll need the coordinates of C and D.
It doesn’t end here. To find the coordinates of C and D, we’ll need those of P, Q, S, and T. And for that, we’ll need the equations of PQ and ST (which need to be solved with the respective circles).
For the equations of PQ and ST, we’ll need the coordinates of R. Then, the standard method to find the equation of a tangent from an external point will give us the equations.
Finally, to find the coordinates of R, we’ll apply simple geometry – the triangles ROP and RAQ are similar. That means RO/RA = OP/AQ or 1/r. In other words, R divides OA in the ratio 1 : r externally.
And using the section formula, we’ll get the coordinates of R as ((4/(1 – r), 1/(1 – r)).
I won’t detail out the solution here, but you can proceed (at your own risk!) in the reverse order to find the value of r².
Method 2
Since C and D are the midpoints of PQ and ST respectively, the line CD must be parallel to both PS and QT (PS and QT being parallel themselves), and equidistant from them. Take a look.
How does that help? We can find the equation of each of PS, QT, and CD and use the above condition to find the value of r.
Let’s start with the equation of CD, which is probably the easiest to find. It passes through B, which lies on the x-axis at a distance of √5 from A(4 , 1). Secondly, due to symmetry, CD will be perpendicular to OB. So, we know its slope as well.
Using the distance formula, B comes out to be (2, 0) (and (6, 0) as well, but we’ll reject that… why?). Using the condition for perpendicular lines, the slope of CD comes out to be -4.
Finally, using the point-slope form, the equation of CD comes out to be y – 0 = -4(x – 2), or 4x + y – 8 = 0.
As a sidenote, using this equation also simplifies the previous method a bit.
Now, PS and QT are the chords of contact from the point R for C1 and C2 respectively. We’d already found the coordinates of R as ((4/(1 – r), 1/(1 – r)).
From this point, the equation of the chord of contact of C1, i.e. the equation of PS is 4x/(1 – r) + y/(1 – r) – 1 = 0, or 4x + y + r – 1 = 0.
Similary, the equation QT is 4x/(1 – r) + y/(1 – r) – 4(x + 4/(1 – r)) – (y + 1/(1 – r)) + 17 – r² = 0. This simplifies to 4x + y + r² – r – 17 = 0.
Let’s rewrite the three equations.
- PS: 4x + y + r – 1 = 0
- CD: 4x + y – 8 = 0
- QT: 4x + y + r² – r – 17 = 0
Since CD is equidistant from the other two, we have [(r – 1) + (r² – r – 17)]/2 = -8. This gives the value of r² as 2. Phew!
Method 3
Since C and D are the midpoints of PQ and ST respectively, CP = CQ and DS = DT.
In other words, tangents drawn from C to C1 and C2 are equal. Similarly, tangents drawn from D to C1 and C2 are equal.
This means C and D lie on the radical axis of the two circles. And so do the midpoints of the other two tangents, which means we could’ve taken any two tangents to begin with.
Now, to proceed we’ll find the equation of the radical axis of the two circles, pass it through B and solve for r.
The equation of the radical axis can be found easily by subtracting the equations of the two circles, when written in standard form, i.e. x² + y² – 1 = 0 and (x – 4)² + (y – 1)² – r² = 0.
On subtracting, we’ll get the required equation as 8x + 2y – 18 + r² = 0.
Finally, on substituting the coordinates of B, we’ll get 8(2) + 2(0) – 18 + r² = 0, which gives r² = 2. Semi phew!
Comments
Needless to say, method 1 is extremely painful and should be avoided, especially in an exam. Method 2 is somewhat less painful and serves as an alternate for method 3, which is the clear winner. If you’re unable to connect the given setting to the concept of radical axis, then the problem becomes quite tedious. Otherwise, it’s a fairly direct application. Still, somewhat heavy on the calculations.