Question Video: Finding the Equation of the Tangent to a Circle with a Given Equation That Forms an Isosceles Triangle with the Positive 𝑥- and 𝑦-Axes Mathematics

A tangent to 𝑥² + 𝑦² = 72 forms an isosceles triangle when taken with the positive 𝑥- and 𝑦-axes. What is the equation of this tangent?

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Video Transcript

A tangent to 𝑥 squared plus 𝑦 squared equals 72 forms an isosceles triangle when taken with the positive 𝑥- and 𝑦-axes. What is the equation of this tangent?

Let’s begin by sketching the curve and the tangent in question. 𝑥 squared plus 𝑦 squared equals 72 is the equation of a circle with its center at the origin and a radius of root 72 units. There is then a tangent to this circle which forms an isosceles triangle with the positive 𝑥- and 𝑦-axes. We know that it is the two sides that lie along the 𝑥- and 𝑦-axes that must be the two equal sides of this isosceles triangle. Because this is also a right triangle and in any right triangle, the hypotenuse is the longest side and is strictly longer than any other side in the triangle, the two shorter sides must therefore be the sides of equal length.

In order to find the equation of a tangent or indeed any straight line, we need to know its slope and the coordinates of at least one point that lies on the line. Let’s begin by considering the slope of this tangent. The slope of the tangent will be the same as the slope of the curve at the point of contact. We don’t know the coordinates of this point, but let’s begin by finding the slope function of the curve. We’ll do this using differentiation. As the curve is defined implicitly, we’ll need to use implicit differentiation when differentiating the second term with respect to 𝑥. Differentiating each term with respect to 𝑥 then gives two 𝑥 plus two 𝑦 d𝑦 by d𝑥 equals zero.

We can make d𝑦 by d𝑥 the subject of this equation by first subtracting two 𝑥 from each side and then dividing by two 𝑦, giving d𝑦 by d𝑥 equals negative two 𝑥 over two 𝑦. Finally, canceling a factor of two gives d𝑦 by d𝑥 equals negative 𝑥 over 𝑦.

So we’ve found an expression for the slope function of the curve, but we don’t have the coordinates of the point where the tangent meets the curve to substitute in. What we do know is that the triangle formed by the tangent and the positive 𝑥- and 𝑦-axes is isosceles. So if the coordinates of the point where the tangent intersects the 𝑥-axis are 𝑎, zero, then the coordinates of the point where the tangent intersects the 𝑦-axis will be zero, 𝑎 so that the two line segments in pink are of equal length.

We can then recall that another way to find the slope of a straight line given two points that lie on the line is using the formula 𝑚 equals 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. Substituting the two points with coordinates zero, 𝑎 and 𝑎, zero and then simplifying gives 𝑚 equals negative one.

So we now know that the slope of the tangent is negative one. In order to find its equation, we need to determine the coordinates of the point that lies on the tangent. At the point where the tangent meets the curve, the slope function of the curve must be equal to the slope of the tangent, which we know to be negative one. So equating these two expressions gives the equation negative 𝑥 over 𝑦 equals negative one. Multiplying both sides of this equation by negative 𝑦 gives 𝑥 equals 𝑦.

We now know that at the point where the tangent meets the curve, the 𝑥- and 𝑦-coordinates are the same. We can therefore replace 𝑦 squared with 𝑥 squared in the equation of the curve to give an equation in 𝑥 only, which will enable us to find the 𝑥-coordinate of this point. Solving this equation, we find that 𝑥 squared is equal to 36. 𝑥 is therefore equal to the square root of 36, taking only the positive solution here as the point lies in the first quadrant, and so we know 𝑥 is positive. We find then that the 𝑥-coordinate of this point is six.

As we’ve determined that the 𝑥- and 𝑦-coordinates are equal at this point, it follows that the 𝑦-coordinate is also equal to six.

We’ve now determined the slope of the tangent and the coordinates of one point that lies on it, and so we’re ready to find its equation. Using the point–slope form of the equation of a straight line and substituting negative one for the slope 𝑚 and the point six, six for 𝑥 one, 𝑦 one, we find that the equation of the line is 𝑦 minus six equals negative one multiplied by 𝑥 minus six. Distributing the negative one over the parentheses on the right-hand side and then collecting all terms on the left-hand side of the equation, we find that the equation of the tangent is 𝑥 plus 𝑦 minus 12 equals zero.