A $4 x^{2}+9 y^{2}=36$ ellipse has tangents drawn from the point $P\left(\frac{3}{5}, \frac{14}{5}\right)$. What is the equation of these tangents? Determine the equation and area of the circle that passes through the point and the points of tangency.
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$

The equation of the tangent to the ellipse at some point $(\xi, \eta)$ is:

$$
\frac{x \xi}{a^{2}}+\frac{y \eta}{b^{2}}=1
$$

If the tangent passes through the point $P\left(\xi_{0}, \eta_{0}\right)$, then:

$$
\frac{\xi_{0} \xi}{a^{2}}+\frac{\eta_{0} \eta}{b^{2}}=1
$$

and since $\xi, \eta$ lies on the ellipse, we have:

$$
\frac{\xi^{2}}{a^{2}}+\frac{\eta^{2}}{b^{2}}=1
$$

From equations (3) and (4), $\xi$ and $\eta$ can be calculated, which makes equation (2) fully determined. Since we generally get two values $\left(\xi_{1}, \xi_{2} ; \eta_{1}, \eta_{2}\right)$, two tangents can usually be drawn from a point to the ellipse. In our case: $a=3 ; b=2 ; \eta_{0}=\frac{3}{5} ; \eta_{q}=\frac{14}{5}$, so equations (3) and (4) become:

$$
\begin{aligned}
2 \xi+21 \eta & =30 \\
4 \xi^{2}+9 \eta^{2} & =36
\end{aligned}
$$

From these equations:

$$
P_{1}\left(\xi_{1}=-\frac{9}{5} ; \eta_{1}=\frac{8}{5}\right) \text { and } P_{2}\left(\xi_{2}=\frac{12}{5} ; \eta_{2}=\frac{6}{5}\right) \text {. }
$$

Now we can get the equations of the tangents $P P_{1}$ and $P P_{2}$ by substituting $\xi_{1}, \eta_{1}$ and $\xi_{2}, \eta_{2}$ into (2):

$$
P P_{1} \equiv \frac{x}{-5}+\frac{2 y}{5}=1
$$

and

$$
P P_{2} \equiv \frac{x}{\frac{15}{4}}+\frac{y}{\frac{10}{3}}=1
$$

What is the equation of the circle that passes through the points $P\left\{\xi_{0}, \eta_{0}\right\} ; P_{1}\left\{\xi_{1}, \eta_{1}\right\} ; P_{2}\left\{\xi_{2}, \eta_{2}\right\}$? The equation of the circle we are looking for will always have the form:

$$
\left(x^{2}+y^{2}\right)+A x+B y+C=0
$$

and since this circle passes through the points $P, P_{1}, P_{2}$, we also have:

$$
\left(\xi_{0}^{2}+\eta_{0}^{2}\right)+A \xi_{0}+B \eta_{0}+C=0
$$

$$
\left(\xi_{1}^{2}+\eta_{1}^{2}\right)+A \xi_{1}+B \eta_{1}+C=0
$$

$$
\left(\xi_{2}^{2}+\eta_{2}^{2}\right)+A \xi_{2}+B \eta_{2}+C=0
$$

These are four homogeneous linear equations in the unknowns $1, A, B, C$. Since not all of the unknowns are zero (since $1 \neq 0$), the determinant of the system must be zero (K. M. L. IX. year, p. 179) or:

![](https://cdn.mathpix.com/cropped/2024_05_02_f678f2e60032af84789cg-1.jpg?height=197&width=262&top_left_y=2363&top_left_x=921)

The curve represented by equation $\varepsilon$, as can be seen from the expansion, is a circle because the coefficients of $x^{2}$ and $y^{2}$ are the same and there is no $x y$ term. It is also easily verifiable that the circle $\varepsilon$ passes through each of the given points, because substituting $\xi_{0} ; \eta_{0}, \xi_{1} ; \eta_{1}$ or $\xi_{2} ; \eta_{2}$ results in two identical rows in the determinant, so its value is zero; for example:

![](https://cdn.mathpix.com/cropped/2024_05_02_f678f2e60032af84789cg-2.jpg?height=188&width=337&top_left_y=161&top_left_x=883)

Thus, equation $(\varepsilon)$ is the equation of the circle passing through the three given points. If $P, P_{1}, P_{2}$ are collinear, then the area of the triangle they form is 0, so:

$$
\left|\begin{array}{lll}
\xi_{0} & \eta_{0} & 1 \\
\xi_{1} & \eta_{1} & 1 \\
\xi_{2} & \eta_{2} & 1
\end{array}\right|=0
$$

In this case, the coefficients of $x^{2}$ and $y^{2}$ are 0, so $\varepsilon$ is not a circle but a line equation. It is clear, therefore, that a real circle can only be drawn through three points if the three points do not lie on a straight line. ( $A$ )

In our case, the equation of the circle is:

$$
\left|\begin{array}{cccc}
x^{2}+y^{2} & x & y & 1 \\
\frac{205}{25} & \frac{3}{5} & \frac{14}{5} & 1 \\
\frac{145}{25} & -\frac{9}{5} & \frac{8}{5} & 1 \\
\frac{180}{25} & \frac{12}{5} & \frac{6}{5} & 1
\end{array}\right|=\frac{1}{5^{3}}\left|\begin{array}{cccc}
x^{2}+y^{2} & x & y & 1 \\
41 & 3 & 14 & 1 \\
29 & -9 & 8 & 1 \\
36 & 12 & 6 & 1
\end{array}\right|=0
$$

Expanding this, we get:

$$
150 x^{2}+150^{2}-66 x-168 y-3600=0
$$

or:

$$
x^{2}+y^{2}-\frac{11}{25} x-\frac{28}{25} y=24
$$

Thus:

$$
\left(x-\frac{11}{50}\right)^{2}+\left(y-\frac{14}{25}\right)^{2}=\left(\frac{11}{50}\right)^{2}+\left(\frac{14}{25}\right)^{2}+24
$$

and thus

$$
r^{2}=\left(\frac{11}{50}\right)^{2}+\left(\frac{28}{50}\right)^{2}+24=\frac{60905}{50^{2}}
$$

so the radius of the circle is:

$$
r=\frac{1}{50} \sqrt{60905} \text { units }
$$

and its area is:

$$
T=r^{2} \pi=\frac{60905}{50^{2}} \pi=76.446 \text { square units. }
$$

(Alfréd Haar, Budapest.)