Question Video: Finding the Equation of a Circle Mathematics

What is the equation of the circle of centre (3, 4) and passing through (7, 7)?


Video Transcript

What is the equation of the circle of centre three, four and passing through seven, seven?

So we have the centre of a circle and the coordinates of one of the points on the circumference, and we’re looking to find the equation of this circle. We’ll answer this question using the standard form for the equation of a circle, which is that if a circle has centre β„Žπ‘˜ and radius π‘Ÿ, then its equation is π‘₯ minus β„Ž all squared plus 𝑦 minus π‘˜ all squared is equal to π‘Ÿ squared.

We already know the values of β„Ž and π‘˜ for the centre of the circle as we’re given them in the question. What we need to do is work out the radius of the circle. Now the radius, remember, is the distance between the centre of the circle and each point on the circumference. So in order find the radius, we need to calculate the distance between the two points three, four and seven, seven. We can do this using the distance formula between two points.

We have that the radius of the circle is equal to the square root of seven minus three all squared plus seven minus four all squared. This tells us that the radius is equal to the square root of four squared plus three squared, and if we work through the next stages here, that’s the square root of 16 plus nine, which is the square root of 25. And as 25 is a square number, the radius is just equal to five.

So now we have both the centre of the circle and its radius, so we can substitute the values of β„Ž, π‘˜, and π‘Ÿ into the equation of the circle. So we have π‘₯ minus three squared plus 𝑦 minus four squared is equal to five squared.

Now if we were asked for our answer in the centre radius form, we would be able to leave it like this or perhaps evaluate five squared as 25, but in this case, we’re gonna give the answer in the expanded form, so we’re going to expand each pair of brackets and then simplify the result.

When I expand π‘₯ minus three all squared, I get π‘₯ squared minus six π‘₯ plus nine. Remember it’s not just π‘₯ squared plus nine or π‘₯ squared minus nine; there’s that negative six π‘₯ term as well. Expanding the second bracket, I have 𝑦 squared minus eight 𝑦 plus 16, and then this is all equal to five squared, which is 25.

Now finally, let’s just simplify this a little. I have plus nine and plus 16 on the left-hand side, which is equal to 25. So in fact, those just cancel out. And now I’m going to reorder the terms so that they’re in a more standard format, which is π‘₯ squared then 𝑦 squared then π‘₯ then 𝑦 and then the constant term, if there is one.

Remember the constant terms in this equation cancelled out, so the equation of the circle with centre three, four passing through the point seven, seven is π‘₯ squared plus 𝑦 squared minus six π‘₯ minus eight 𝑦 is equal to zero.

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