Question Video: Using the Power of a Point Theorem for Secant Segments to Find the Power of a Point Mathematics • First Year of Secondary School

Video Transcript

Two circles 𝑀 and 𝑁 intersect at points 𝐴 and 𝐵, and the point 𝐶 satisfies 𝐶 belongs on the line 𝐵𝐴 and 𝐶 does not belong on the line segment 𝐵𝐴. 𝐷 and 𝐸 are the points where the line segment 𝐶𝐸 intercects the circle 𝑀, and the line 𝐶𝐹 is a tangent to 𝑁. Given that 𝐶𝐷 equals seven and 𝐷𝐸 equals 12, find 𝑃 sub 𝑁 of 𝐶.

There’s an awful lot of information given to us in the question, so let’s begin by drawing a diagram. We have two circles with centers 𝑀 and 𝑁. We don’t know which is larger. And actually, it doesn’t really matter. These two circles intersect at the points 𝐴 and 𝐵. Now we’re told that the point 𝐶 is on the line 𝐵𝐴, but it isn’t on the line segment 𝐵𝐴. That means if we draw in the line 𝐵𝐴, 𝐶 is somewhere on this line, but it isn’t between 𝐴 and 𝐵. So perhaps it’s here. There’s then a line segment 𝐶𝐸, which intersects the circle 𝑀 at points 𝐷 and 𝐸, so we can add this line segment to our diagram. There is then a line 𝐶𝐹 which is a tangent to circle 𝑁. So, here’s that final line. The last information we’re given is the lengths of two line segments. 𝐶𝐷 is seven units and 𝐷𝐸 is 12 units.

So, we have a diagram. And now let’s look at what we’re being asked to find. 𝑃 sub 𝑁 of 𝐶 means the power of point 𝐶 with respect to circle 𝑁. It’s calculated using the formula 𝐶𝑁 squared minus 𝑟 squared, where 𝑟 is the radius of the circle 𝑁. So it’s the distance between points 𝐶 and the center of the circle squared minus the radius squared. However, we don’t have any of this information, so we’re going to need a different approach. Let’s consider circle 𝑀 first as we have more information about this circle. In circle 𝑀, we know the lengths of 𝐶𝐷 and 𝐷𝐸, which are each segments of a secant to this circle. We can therefore recall the power of a point theorem concerning the lengths of secant segments. This states the following.

Consider a circle 𝑀 and a point 𝐶 outside the circle. Let the line segment 𝐶𝐸 be a secant segment to the circle at 𝐷 and 𝐸. Then the power of point 𝐶 with respect to circle 𝑀 is equal to 𝐶𝐷 multiplied by 𝐶𝐸. This is great because we know the length of 𝐶𝐷. It’s seven units. And the length of 𝐶𝐸 is seven plus 12. It’s 19 units. So we can work out the power of point 𝐶 with respect to circle 𝑀 as seven multiplied by 19, which is 133.

This isn’t what we were asked to find though. We were asked to find the power of point 𝐶 with respect to circle 𝑁. Well, the line at 𝐶𝐵 or 𝐵𝐶 is also a secant of the circle 𝑀, so it follows that the product of the lengths of the secant segments 𝐶𝐴 and 𝐶𝐵 will also be equal to 133. This is because the power of a point with respect to any given circle is always the same. So we have the equation 𝐶𝐴 multiplied by 𝐶𝐵 is 133. But this is useful because 𝐶𝐵 isn’t just a secant segment of circle 𝑀; it’s also a secant segment of circle 𝑁. It is, in fact, a common secant segment for the two circles.

So by the power of a point theorem concerning the length of secant segments for circle 𝑁, we have that 𝑃 𝑁 of 𝐶 is equal to 𝐶𝐴 multiplied by 𝐶𝐵. And we’ve just determined that this will be equal to 133. In other words, what we’ve found then is that the power of point 𝐶 with respect to each circle found using their common secant is the same. We can conclude then that the power of point 𝐶 with respect to circle 𝑁 is 133.