Video Transcript
Consider the circle 𝑥 minus five squared plus 𝑦 plus two squared is equal to 25. Is the line 𝑦 minus three equals zero tangent to, intersecting, or disjoint from the circle?
So we’ve been given the equation of a circle and the equation of a line and we have three possibilities for the relationship between the line and the circle. We’ll look at these three possibilities in a bit more detail in a moment, but first let’s consider the equation of this line.
The line 𝑦 minus three equals zero is equivalent to the line 𝑦 equals three and you can see this by adding three to both sides of this equation. As this line has the equation 𝑦 equals a constant, it is in fact the horizontal line. Now let’s think about those three possible scenarios: intersecting, disjoint, or tangent.
So suppose this is the circle in question. The first possibility is that the horizontal line 𝑦 equals three is entirely above or perhaps entirely below the circle and therefore it is disjoint. The second possibility is that this line passes through the circle and therefore there are two points of intersection between the circle and the line. The third and final possibility is that the line doesn’t actually pass inside the circle, but it just touches the circumference at one point and is therefore a tangent to the circle.
So this is the physical interpretation of the three possibilities given in the question. Now let’s think about how we’re going to determine which of these is the case. Well, it’s all linked to the number of points of intersection between the line and the circle or in other way, it’s linked to the number of solutions when we solve the equations of the line and the circle simultaneously. If there are no solutions to the simultaneous equations, then the line and the circular disjoint, if there’s one solution, then the line is a tangent to the circle, and if there are two separate solutions, then the line intersects the circle.
So by solving the simultaneous equations, we’ll be able to see which of these three situations we’re in. So let’s look at solving these simultaneous equations. We’ll use the rearranged version of the equation of the line: 𝑦 is equal to three. The most straightforward way to solve these simultaneous equations is going to be to substitute from the equation of the line into the equation of the circle.
So we’ll substitute three for 𝑦. So we have 𝑥 minus five all squared plus three plus two all squared is equal to 25. Now, three plus two is five and five squared is 25. So we have a 25 on both sides of the equation, which would just cancel each other out directly. This then leaves 𝑥 minus five all squared is equal to zero. Next we want to take the square root of both sides of this equation. So we have the 𝑥 minus five is equal to zero.
Finally, to solve this equation for 𝑥, we need to add five to both sides. So we have 𝑥 is equal to five. Remember the equation of our line was 𝑦 equals three, and so that’s the solution for 𝑦. So we have one pair of values for 𝑥 and 𝑦. 𝑥 is equal to five and 𝑦 is equal to three.
This means we have one solution to the simultaneous equations. It also means that the line and the circle intersect at only one point, the point with coordinates five, three. Therefore, our answer to the question as to which of these three situations exist is that the line is a tangent to the circle.
