Find the equations of the two tangents to the circle 𝑥 squared plus 𝑦 squared equals 125 that are inclined to the positive 𝑥-axis by an angle whose tangent is two.
Let’s think about what we actually know here. The circle with equation 𝑥 squared plus 𝑦 squared equals 125 has a center at the origin zero, zero. What’s not entirely necessary to know its radius is equal to the square root of 125, or five root five units. What we’re actually interested in is finding the equations of the two tangents to the circle inclined to the positive 𝑥-axis by an angle whose tangent is two. Now these might look a little something like this.
But what does it mean if they’re inclined to the positive 𝑥-axis by an angle whose tangent is two? Well, let’s suppose the angle is equal to 𝜃. Thinking about the right triangle, we know that the tangent ratio links the opposite side in this triangle to the adjacent. Specifically, tan 𝜃 is equal to opposite over adjacent. This means the ratio of the opposite to the adjacent side is equal to two. We could then say that the opposite side is equal to two length units and the adjacent side is equal to one. Then the length of the opposite divided by the adjacent side is two as we required.
Of course, we could have had any multiple of these values, but we’ll see why this is the best option in a moment. Now, since the two tangents are straight lines, we know we can use the formula 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one to find their equations. Now here 𝑚 is the slope of the line, whilst 𝑥 one, 𝑦 one is a point that they pass through. But of course, 𝑚, the slope, can be found by calculating rise over run or, equivalently, change in 𝑦 divided by change in 𝑥. Now, in this case, we can see that the change in 𝑦 is two and the change in 𝑥 is one. So the slope of the line is two divided by one which is equal to two.
Now, it’s no coincidence that the slope of our tangents are equal to the value of the tangent of the angle at which they’re inclined by. In general, the slope of line is equivalent to the tangent of the angle of the slope. So we actually have the slope of both of our lines; they’re two. But we’re going to need to find the points that they pass through. And so we’re going to go back to the equation of the circle; that is, 𝑥 squared plus 𝑦 squared equals 125.
We know that given an equation of a curve, we can find the slope of the curve at a given point by thinking about the derivative. So what we’re going to do is find the derivative function for the equation of our circle and then set this equal to two. This will tell us the points at which our tangents touch the circle. So how do we differentiate the equation 𝑥 squared plus 𝑦 squared equals 125? We’re going to do it in fact term by term. First, let’s differentiate 𝑥 squared with respect to 𝑥. Using the general power rule for differentiation, we multiply by the exponent and then reduce that exponent by one. So its derivative is simply two 𝑥.
But what about the derivative of 𝑦 squared with respect to 𝑥? We use implicit differentiation here. This is a special version of the chain rule. Essentially, we differentiate with respect to 𝑦 and then multiply by d𝑦 by d𝑥. The derivative of 𝑦 squared with respect to 𝑦 is two 𝑦. So when we differentiate 𝑦 squared with respect to 𝑥, we get two 𝑦 times d𝑦 by d𝑥. Then, on the right-hand side, the derivative of 125 with respect to 𝑥 is equal to zero. So our expression for the derivative is two 𝑥 plus two 𝑦 d𝑦 by d𝑥 equals zero.
Let’s divide through by two and then make d𝑦 by d𝑥 the subject. We can subtract 𝑥 from both sides, and then we divide through by 𝑦. And that gives us that d𝑦 by d𝑥 is equal to negative 𝑥 over 𝑦. We want to find the points where the slope is equal to two because that’s the slope of the tangents we’re interested in. So we set d𝑦 by d𝑥 equal to two. So two is negative 𝑥 over 𝑦. And if we rearrange to make 𝑥 the subject, multiplying by 𝑦 and then multiplying by negative one, we get negative two 𝑦 equals 𝑥.
This shows us the relationship between the 𝑥- and 𝑦-values at the two points where our lines meet the circle. This means then that we can substitute this into the equation for the circle and we’ll find the two points where they meet. Replacing 𝑥 with negative two 𝑦 and our equation becomes negative two 𝑦 squared plus 𝑦 squared equals 125. Negative two 𝑦 squared is four 𝑦 squared, so the left-hand side becomes five 𝑦 squared. Then, we divide through by five. So 𝑦 squared is equal to 25.
Our final step will be to take the square root. Now, we need to think about both the positive and negative square root of 25. Since the square root of 25 is five, we can say that 𝑦 must be equal to positive or negative five. Then, we can substitute this back into the equation 𝑥 equals negative two 𝑦 to find the corresponding 𝑥-values. When we do, we find that 𝑥 is equal to negative or positive 10. In other words, our two tangents pass through the points 10, negative five and negative 10, five.
We now have the two points through which our two tangents pass and the value of their slope. So we can substitute all of this into our equation for a straight line. Beginning with the point 10, negative five — that will be this point here — we get 𝑦 minus negative five equals two times 𝑥 minus 10. Distributing the parentheses, this becomes 𝑦 plus five equals two 𝑥 minus 20. And then subtracting two 𝑥 and adding 20, we find the equation of our first line is 𝑦 minus two 𝑥 plus 25 equals zero.
Let’s now repeat this with the point negative 10, five; that’s this one. Substituting in, we get 𝑦 minus five equals two times 𝑥 minus negative 10. That gives us 𝑦 minus five equals two 𝑥 plus 20. And rearranging once again and we find the equation of this line is 𝑦 minus two 𝑥 minus 25 equals zero. And so our equations are 𝑦 minus two 𝑥 plus 25 equals zero and 𝑦 minus two 𝑥 minus 25 equals zero.