19. A circle $C$, of radius $r$, passes through the points $A (a, 0)$, $A_{1} (-a, 0)$ and $B (0, b)$, where $a$ and $b$ are positive and are not equal; a circle $C_{1}$, of radius $r_{1}$, passes through $A, B$ and $B_{1} (0, -b)$. Prove that the centre of $C$ is $\displaystyle\left(0, \frac{b^{2} - a^{2}}{2b}\right)$ and that $r_{1}/r = b/a$.
Find the point of intersection of the tangents to the circle $C$ at $A$ and $A_{1}$.
I have completed the two proofs, but have gone wrong with the final part of the question.
I have said:
The centre of $C$ is $\left(0, \dfrac{b^{2} - a^{2}}{2b}\right)$.
The gradient of the radius to $A$ is:
$$\left.\left(\dfrac{b^{2} - a^{2}}{2b} - 0\right)\right/(0 - a) = \dfrac{b^{2} - a^{2}}{-2ab} = \dfrac{a^{2} - b^{2}}{2ab}.$$
$\therefore$ The equation of the tangent to $C$ at $A$ is:
$$y = -\dfrac{2ab}{a^{2} - b^{2}}(x - a).$$
The gradient of the radius to $A_{1}$ is: $\dfrac{b^{2} - a^{2}}{2ab}$.
$\therefore$ The equation of the tangent to $C$ at $A_{1}$ is:
$$y = -\dfrac{2ab}{b^{2} - a^{2}}(x + a).$$
$\therefore$ At the point of intersection:
$y = -\dfrac{2ab}{a^{2} - b^{2}}(x - a) \qquad (1)$
$y = -\dfrac{2ab}{b^{2} - a^{2}}(x + a) \qquad (2)$
I chose to eliminate $y$:
$-\dfrac{2ab}{a^{2} - b^{2}}(x - a) = -\dfrac{2ab}{b^{2} - a^{2}}(x + a)$
$(x - a)(b^{2} - a^{2}) = (x + a)(a^{2} - b^{2})$
$xb^{2} - xa^{2} - ab^{2} + a^{3} = xa^{2} - xb^{2} + a^{3} - ab^{2}$
$2xb^{2} - 2xa^{2} = 0$
$x = 0$
Sub $x$ into (2)
$y = -\dfrac{2a^{2}b}{b^{2} - a^{2}}$
But, the book of knowledge saith:
$(0, 2a^{2}b/(b^{2} - a^{2})).$
Which presumably means I've made a mistake with a sign somewhere (but may indicate a more significant misunderstanding); but I'm afraid I simply can't spot what I've done wrong - any ideas?
$\endgroup$ 12 Answers
$\begingroup$Compare the above drawings and determine who's answer is more correct.
$\endgroup$ 1 $\begingroup$You made a sign error quite early on when you used the gradient at $A_1$, but the $x$-coordinate at $A$ to form the equation of one of the tangents. (Or perhaps you mistakenly used $x+x_0$ instead of $x-x_0$.) The correct equation for the tangent at $A_1$ is $$y=-{2ab\over a^2-b^2}(x\color{red}+a).$$
There wasn’t really any need to go through the rest of the calculations in your question in order to come up with the solution, though. By symmetry, we already know that the intersection of the tangents is on the $y$-axis, if for no other reason than the fact that it must lie on the perpendicular bisector of the chord $\overline{AA'}$. Once you have an equation for one of the tangents, you can immediately plug in $x=0$ into it.
Alternatively, you can find the intersection of the tangents without constructing explicit tangent line equations. The point of intersection, the circle’s center and $A$ form a right triangle, so a straightforward application of the Pythagorean theorem will get you the intersection point. For this problem, your approach requires much less work, though.
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