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A circular fountain has the equation x² + y² - 2x + 4y - 20 = 0, with lengths in metres…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A circular fountain has the equation $x^2 + y^2 - 2x + 4y - 20 = 0$ , with lengths in metres. A water jet leaves the rim at the point $(4, 2)$ and goes along the tangent to the circle there. A support rod goes along the normal at the same point. Find the slope of the tangent, the slope of the normal, and the $x$ -intercept of the tangent.

Need a reminder? The key ideaShow

A point is inside, on or outside a circle as $S_1 < 0$ , $= 0$ or $> 0$ . A line at distance $d$ from the centre cuts the circle in a chord of length $2\sqrt{r^2 - d^2}$ if $d < r$ . The tangent at $(x_1, y_1)$ is $T = 0$ and the normal passes through the centre.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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