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Question

Hi, I have Just gone into the Easter holidays and I am doing a lot a maths revision. I have been met with this question in one of my practice papers, here is the question:

A circle has equation x^2 + y^2 = 34. P lies on the circle and has x-coordinate 3. The tangent at P intersects the x-axis at point A and the y-axis at point B.

Image: https://imgur.com/a/aFgbX

Work out the co-ordinates of A and B.

Disclaimer

I am not really interested in the answer to the question but I was quite confused by this question as I am not sure how I can get the tangents length through algebra (so for general use rather than specific question). If anyone has any ideas about how you can easily find the length of a tangent, please let me know!

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I assume you know (or once knew) enough algebra and analytic geometry to:

  • Write the equation of the circle.

  • Find the $y$ coordinate of the intersection.

  • Find the slope of the tangent there (since the tangent is perpendicular to the radius)

  • Find the equation of the tangent line.

  • Find the $x$ and $y$ intercepts for that line.

    The length of the tangent is a red herring. You don't need it to solve the problem. Once you have the solution it's easy to compute.

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From Geometry: We know that the radial line is always perpendicular to the tangent line of a circle, so $9+6^2=34 \implies y= \pm5$. Assume $y=5$. Then the tangent line has slope $-3/5$, and it goes through the point $(3,5)$, so you can solve $y=-3/5(x)+b$ to get the $y$-intercept: $5=-9/5+b \implies 34/5=b$, which is the $y$-intercept and the $x$-intercept is similar.

From Calculus: $x^2+y^2=34$ implies that $2x +2y \frac{dy}{dx}=0$, so $\frac{dy}{dx}=-x/y$, which is negative reciprocal to the slope, so you can do as before.

Here is the general idea: a line is characterized by a slope and a co-ordinate, so you first solve for the co-ordinate and guess the slope. Once you have the equation for a line, you can solve for the intercepts by individually setting $x=0$ and $y=0$. You don't need to solve for the distance.

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