Video Transcript
A circle has center four, negative two and goes through the point negative two, negative three. Find the equation of the circle.
To answer this question, we can recall the center-radius form of the equation of a circle. If a circle has center with coordinates ℎ, 𝑘 and a radius of 𝑟 units, then its equation can be written in center-radius form: 𝑥 minus ℎ squared plus 𝑦 minus 𝑘 squared equals 𝑟 squared.
We’ve been given the center of our circle in the question; it’s four, negative two. So the value of ℎ is four and the value of 𝑘 is negative two. Substituting these values into the center-radius form gives 𝑥 minus four squared plus 𝑦 minus negative two squared equals 𝑟 squared.
In the second bracket, the two negative signs next to each other form a positive. 𝑦 minus negative two is the same as 𝑦 plus two. So our equation simplifies to 𝑥 minus four squared plus 𝑦 plus two squared equals 𝑟 squared.
To complete the question then, we need to find the radius of our circle. Now, the radius of a circle is the distance between its center and every point on its circumference. We know the center of our circle. And we’re also told that the circle goes through the point negative two, negative three. So this point is on the circumference.
We need to recall then the distance formula, which tells us how to calculate the distance between two points on a coordinate grid. If those points have coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, then the distance between them can be found by finding the square root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared.
This looks complicated. But actually, it doesn’t matter which way around you label the point as 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two. All this formula is saying is we need to find the difference between the 𝑥-coordinates and the difference between the 𝑦-coordinates, square each of these values, add them together, and then take the square root.
Now, actually, we can make this a little bit simpler as this will tell us the distance between the points which is the radius of the circle. But in the equation of our circle, we actually use 𝑟 squared. So rather than working out 𝑑, which is the square root of this long formula, we can work out 𝑑 squared, which means we don’t need to take the square root.
I’ve chosen to make the center of the circle the point 𝑥 two, 𝑦 two and the point on the circumference 𝑥 one, 𝑦 one. But it doesn’t matter which way around you do this. So we have the 𝑟 squared is equal to four minus negative two squared, that’s for 𝑥 two minus 𝑥 one squared, plus negative two minus negative three squared, that’s for 𝑦 two minus 𝑦 one squared.
Now, four minus negative two is the same as four plus two which is six and negative two minus negative three is the same as negative two plus three which is one. So we have six squared plus one squared. Six squared is 36 and one squared is one. So we have 𝑟 squared is equal to 36 plus one which is 37.
We can then substitute this value of 𝑟 squared into the equation of our circle. The equation of the circle then is 𝑥 minus four squared plus 𝑦 plus two squared equals 37. This equation is in center-radius form. And as we haven’t been asked to give our answer in a particular format, there is no need to expand it.
