# Assignment 23, Q3(ii)

I noticed that the triangle is a right-angled triangle and remembered a result from the warm down of assignment 21 about incircles of right-angled triangles, namely that there's a square of side length $r$ (the radius of the incircle) in the corner of the triangle. I found the length of the hypotenuse along the $x$-axis to be 5, and in part (i) I had noticed that the circle is tangent to the $x$-axis at $x=2t$.

I could then use Pythag to relate all the sides of my triangle, now in terms of $t$, and obtained $5^2=(5-2t+t)^2+(t+2t)^2$ Which solves out easily to give $t=1$.

Keep up the assignments. They're great fun.

### Nice!

This is a nice alternative way of doing this, thanks!

### Another similar way

Setting $(x_0, y_0)$ to be the point where the incircle touches the line $4y+3x=15$, from the equation of the circle we can derive the slope of the tangent at $(x_0, y_0)$, which is $\frac{dy}{dx}=\frac{2t-x_0}{y_0-t}$, thus obtaining the equation of the tangent line: $y-y_0=(\frac{2t-x_0}{y_0-t}) (x-x_0)$ which rearranges to $y(y_0-t)+x(x_0-2t)=y_0(y_0-t)-x_0(2t-x_0)$. This line is the same as $4y+3x=15$ $(*)$ so the ratios of the $x$ and $y$ coefficients must be the same $\Longrightarrow$ $\frac{y_0-t}{4}=\frac{x_0-2t}{3}$ and plugging back into $(*)$ we obtain: $x_0=\frac{9+4t}{5}$, $y_0=\frac{12-3t}{5}$. Finally, substituting these values into the equation of the circle, we get the rather nice quadratic $3t^2-12t+9=0$ which solves to get $t=1$ (and $t=3$ for the excircle).

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