Question Video: Recognising Properties of Circle Construction | Nagwa Question Video: Recognising Properties of Circle Construction | Nagwa

# Question Video: Recognising Properties of Circle Construction Mathematics • Third Year of Preparatory School

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Consider the two points π΄ and π΅. What is the radius of the smallest circle that can be drawn in order to pass through the two points?

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### Video Transcript

Consider the two points π΄ and π΅. What is the radius of the smallest circle that can be drawn in order to pass through the two points?

In this question, we need to determine the radius of the smallest circle which passes through two points π΄ and π΅. To answer this question, we start by recalling that every point on a circle will be equidistant from its center. And we also recall another fact. Every point which is equidistant from two points π΄ and π΅ will lie on the perpendicular bisector of π΄ and π΅. So the center of our circle must lie on the perpendicular bisector of π΄ and π΅. So letβs start by constructing this. We need to draw the line segment from π΄ to π΅ and then mark the midpoint of this line segment. Weβll call this π. Then the perpendicular bisector is the line perpendicular to π΄π΅ which passes through π. Any point which lies on this line is equidistant from π΄ and π΅.

And so these can be choices for the center of our circle. For example, π, π two, π three, or π four could be the centers of our circle. Then the radius of this circle would just be the distance between the center and any point on the circle, for example, the distance between the center and point π΄. Graphically, it appears the further we get away from the point π, the larger this radius is. In fact, we can prove this is true. For example, we can see that triangle π΄ππ two is a right triangle. In particular, this means its hypotenuse must be longer than the other sides. π΄ππ three is also a right triangle, and π΄ππ four is another right triangle. So π three is bigger than π, and π four is bigger than π. Therefore, π will be the circle with the smallest radius which passes through π΄ and π΅.

And although we were not asked to in this question, we can even sketch this circle by putting the points of our compass at the point π and then the tip of our compass at either the points π΄ or π΅. We were asked to find the radius of this circle. Remember, we chose π to be the midpoint of the line segment π΄π΅. And this means the value of π is going to be one-half the distance from π΄ to π΅. In other words, the radius of the smallest circle which passes through two distinct points π΄ and π΅ is one-half the distance from π΄ to π΅.

So far, weβve constructed circles through one point and through two distinct points. Now letβs try to construct the circle through three distinct points π΄, π΅, and πΆ. To do this, we need to find the center of our circle which needs to be equidistant from all three points. If the center of the circle is equidistant from all three points, then it must be equidistant from both π΄ and π΅. And we can find the set of all points equidistant from both π΄ and π΅ by finding the perpendicular bisector. So we start by finding the midpoint of the line segment between π΄ and π΅. And then we draw the line through the midpoint π one which is perpendicular to the line segment π΄π΅. Every point equidistant from both π΄ and π΅ lies on the perpendicular bisector between π΄ and π΅. So if the center of our circle exists, it must lie on this line.

But the exact same reasoning is true for points πΆ and π΅. The center of our circle is equidistant from πΆ and π΅, so it must lie on the perpendicular bisector of the line segment π΅πΆ. So weβll also sketch the perpendicular bisector of line segment π΅πΆ. We find the point π two which is the midpoint of this line segment and then draw a line perpendicular to this line which passes through π two. Every point on this line is equidistant from π΅ and πΆ, and we can see thereβs a point of intersection between the two perpendicular bisectors. Since this point is on both perpendicular bisectors, itβs equidistant from π΄, π΅, and πΆ. This means we can construct a circle with center π which passes through all three points π΄, π΅, and πΆ. The radius of this circle will then be the distance between either π and π΄, π and π΅, or π and πΆ.

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