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?
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