Video: Finding the Equation of 𝐴 Circle in the Coordinate System

Determine the general equation of the shown circle 𝑀 passing through the origin point and the two points 𝐴(8, 0) and 𝐡(0, βˆ’10).

08:27

Video Transcript

Describe the general equation of the shown circle 𝑀 passing through the origin point and the two points 𝐴: eight, zero and 𝐡: zero, negative 10.

So we’ve been asked to find the equation of this circle 𝑀. And we know three points that lie on its circumference, the origin point zero, zero and the two points 𝐴 and 𝐡 whose coordinates have been given. In order to find the general equation of this circle, we’ll first find it in center–radius form. For which we need to know the center of the circle that has the coordinates of the point 𝑀 and also the radius. That would be the length of any of the lines 𝑀𝐴, 𝑀𝑂, or 𝑀𝐡.

We’ll recall, first of all, that the perpendicular bisector of any chord passes through the center of the circle. The lines 𝑂𝐴 and 𝑂𝐡 are each chords of this circle. So if we can find the perpendicular bisectors of these two lines, they’ll meet in the center of the circle, the point 𝑀. Let’s consider the perpendicular bisector of the line 𝑂𝐴 first of all. Remember, perpendicular bisector is a line which divides another line exactly in half. That’s what we mean by bisect. And the perpendicular part means that it meets the line at right angles.

The line 𝑂𝐴 is just a horizontal line. It lies along the π‘₯-axis. It connects the origin, the point zero, zero, with 𝐴, the point eight, zero. As such, its perpendicular bisector will be a vertical line passing through the midpoint of 𝑂𝐴. The π‘₯-coordinate of the midpoint of 𝑂𝐴 will just be the average of the π‘₯-coordinates of the origin and 𝐴. That’s the average of zero and eight, which is four. The 𝑦-coordinate will be zero as this point is on the π‘₯-axis. As the perpendicular bisector of 𝑂𝐴 is a vertical line passing through four on the π‘₯-axis, its equation is therefore π‘₯ equals four.

We can apply similar logic to work out the equation of the perpendicular bisector of 𝑂𝐡. 𝐡 is the point with coordinates zero, negative 10, which means the line 𝑂𝐡 is a vertical line. It lies along the 𝑦-axis. The perpendicular bisector of 𝑂𝐡 will be a horizontal line passing through the midpoint of 𝑂𝐡. The π‘₯-coordinate of this point will be zero as we’re on the 𝑦-axis. And the 𝑦-coordinate will be the average of the 𝑦-coordinates of the origin and the point 𝐡. That’s the average of zero and negative 10, which is negative five.

Therefore, the equation of the perpendicular bisector of 𝑂𝐡 is 𝑦 equals negative five because it is a horizontal line passing through negative five on the 𝑦-axis. The equations of these two lines also give the coordinates of the point 𝑀 as, remember, the point 𝑀 is on both lines. The π‘₯-coordinate of 𝑀 will be four. And the 𝑦-coordinate of 𝑀 will be negative five. So we now know that the center of our circle is the point with coordinates four, negative five.

But what about the radius? Well, the radius is the distance from the center of the circle to any point on the circumference. So, as we said before, we could find the length of either 𝑀𝐴, 𝑀𝑂, or 𝑀𝐡. It will be most straightforward to work out the length of 𝑀𝑂 as the coordinates of the point 𝑂 are zero, zero. To find the distance 𝑑 between two points on a coordinate grid, we use an application of the Pythagorean theorem. If the two points have coordinates π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two, then the distance squared is equal to π‘₯ two minus π‘₯ one squared plus 𝑦 two minus 𝑦 one squared.

So, using the points four, negative five and zero, zero, that’s 𝑀 and the origin, we have that the radius squared is equal to four minus zero squared plus negative five minus zero squared. That’s just equal to four squared plus negative five squared. Four squared is 16. And negative five squared is 25. Remember, a negative multiplied by another negative gives a positive. 16 plus 25 is 41. So we have that the radius of our circle squared is equal to 41. Now, we could square root both sides of this equation to find out what the radius of the circle is equal to rather than the radius squared. But we actually don’t need to because, in the next stage of our working, we’re going to be using the radius squared.

Next, we’ll recall the center–radius form of the equation of a circle. If a circle has its center at the point with coordinates β„Ž, π‘˜ and a radius of π‘Ÿ units, then its equation in center–radius form is given by π‘₯ minus β„Ž all squared plus 𝑦 minus π‘˜ all squared equals π‘Ÿ squared. We know the values of β„Ž and π‘˜. Remember, our circle has a center at the point four, negative five. So β„Ž is four. And π‘˜ is negative five. We’ve also found that π‘Ÿ squared is equal to 41. So we can substitute each of these values into the center–radius form of the equation of a circle.

I’m going to delete some of the working on the screen to make space to do this. But make sure you’ve got it all written down. So we substitute four for β„Ž, negative five for π‘˜, and 41 for π‘Ÿ squared. And we have that the equation of the circle is π‘₯ minus four squared plus 𝑦 minus negative five squared equals 41. In the second bracket, 𝑦 minus negative five is the same as 𝑦 plus five. So our equation becomes π‘₯ miuns four squared plus 𝑦 plus five squared equals 41.

This is the equation of the circle in center–radius form. But we’ve been asked for the general equation, which means we need to expand the brackets and simplify the result. Let’s think about expanding these brackets. Remember, π‘₯ minus four all squared means π‘₯ minus four multiplied by π‘₯ minus four. We must make sure we multiply each term in the first bracket by each term in the second. We don’t just square each term. We can use a method such as the FOIL method to help with our expansion.

First, we multiply the first terms in the two brackets together. That’s π‘₯ multiplied by π‘₯ which gives π‘₯ squared. Then, we multiply the outer terms together. That’s the π‘₯ in the first bracket and the negative four in the second, giving negative four π‘₯. Then, we multiply the inner terms together. That’s the negative four in the first bracket and the π‘₯ in the second, giving another lot of negative four π‘₯. And, finally, we multiply the last terms in the brackets together. Negative four multiplied by negative four is 16.

We can simplify by grouping together the like terms in the center of our expansion. Negative four π‘₯ minus four π‘₯ is equal to negative eight π‘₯. So we have that π‘₯ minus four all squared is equal to π‘₯ squared minus eight π‘₯ plus 16. You can use the same method to show that the expansion of 𝑦 plus five squared is equal to 𝑦 squared plus 10𝑦 plus 25. So substituting the expansions for the two brackets gives π‘₯ squared minus eight π‘₯ plus 16 plus 𝑦 squared plus 10𝑦 plus 25 equals 41.

Now, we can do some simplification because on the left-hand side we have plus 16 plus 25 which is equal to 41. This will cancel with the positive 41 on the right-hand side, which we can also see by subtracting 41 from each side of the equation. When we’re given the equation of a circle in its general form, it’s usual to write the terms in a particular order. We start with the π‘₯ squared terms, then the 𝑦 squared, then the π‘₯s, then the 𝑦s, and then the constant term if there is one.

So, reordering the terms into this standard order gives the equation of our circle as π‘₯ squared plus 𝑦 squared minus eight π‘₯ plus 10𝑦 is equal to zero. Remember, we have no constant term because the positive 16 and positive 25 on the left-hand side combine to cancel with the 41 on the right-hand side. So, we’ve found that the general equation of the circle shown, which passes through the origin and the points eight, zero and zero, negative 10, is π‘₯ squared plus 𝑦 squared minus eight π‘₯ plus 10𝑦 equals zero.

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