### Video Transcript

Which of the following is a point of intersection between circles 𝐴 and 𝐵 where circle 𝐴 has radius 10 and center nine, three and circle 𝐵 has radius five and center zero, seven? (A) zero, seven; (B) three, 11; (C) two, negative one; (D) zero, two; or (E) four, 12.

Well, the first thing we need to do is remind ourselves how we’d find the equation of a circle cause this is what’s gonna help us solve the problem. Well, the general form for the equation of a circle is if we got 𝑥 minus ℎ all squared plus 𝑦 minus 𝑘 all squared, this is equal to 𝑟 squared where the center is ℎ, 𝑘 and 𝑟 is our radius.

Okay, so let’s try and find out what the equation of circle 𝐴 and circle 𝐵 are. Well, for circle 𝐴, what we have is a radius of 10 and a center at nine, three. So therefore, we can say that the equation for circle 𝐴 is gonna be 𝑥 minus nine all squared, because our nine is our ℎ, plus 𝑦 minus three all squared, that’s because our 𝑘 or our 𝑦-coordinate is three, is equal to 100, and that’s because it’s equal to 𝑟 squared. And if the radius is 10, 10 squared is 100.

Well, for circle 𝐵, what we have is that the radius is equal to five and the center is at zero, seven. So therefore, the equation of circle 𝐵, well, we could start with 𝑥 minus zero all squared. However, we don’t need the zero. So therefore, we’re gonna have 𝑥 squared. And this is plus 𝑦 minus seven all squared. And then this is equal to 25. That’s because the radius is five. So therefore, 𝑟 squared is equal to 25. So now, what we need to do is determine which one of our points (A), (B), (C), (D), and (E) are, in fact, a point of intersection between circles 𝐴 and 𝐵. And to do that, what we’re gonna do is substitute the values in to both of our equations to see if it satisfies them both.

So now, if we start with point (A), we know that 𝑥 is equal to zero and 𝑦 is equal to seven. So let’s substitute this into our equations. So if we substitute into the left-hand side for circle 𝐴, we’ll have zero minus nine all squared plus seven minus three all squared, which is gonna give us negative nine squared plus four squared, which is gonna give us 81 plus 16, which is equal to 97. Well, 97 is not equal to 100. So therefore, we can say that point (A) can be ruled out because it can’t be a point of intersection because it doesn’t even lie on circle 𝐴. So therefore, it can’t be a point of intersection that would lie on both of our circles.

So now, what we’re gonna do is move on to point (B). Well, for point (B), we’ve got our 𝑥-coordinate is equal to three and the 𝑦-coordinate is equal to 11. So once again, we can substitute these into the left-hand side for circle 𝐴 equation. And when we’ve done that, we get three minus nine all squared plus 11 minus three all squared. Well, this is gonna give us negative six all squared plus eight squared, which is gonna give us 36 plus 64. And that’s cause if we have a negative multiplied by negative, we get a positive. Well, this is equal to 100. So therefore, it satisfied circle 𝐴. So we know that this point lies on circle 𝐴. So now, let’s see if it lies on circle 𝐵 and, therefore, would be a point of intersection.

Well, if we substitute 𝑥 equals three and 𝑦 equals 11 into the left-hand side of the equation for circle 𝐵, we’re gonna get three squared plus 11 minus seven squared. Well, this’s gonna be equal to nine plus 16 because 11 minus seven is gonna give us four. And four squared is 16. So now, we’ve got nine plus 16, which is equal to 25, which was the right-hand side of the equation. So therefore, we know the point also lies on the equation for circle 𝐵, so it lies on the circle 𝐵. So therefore, we can say that point (B) is the point of intersection between circles 𝐴 and 𝐵. Well, we can see that this is definitely a point of intersection, but let’s check the final three points to see if they are as well.

Well, if we look at point (C), we can see that 𝑥 equals two and 𝑦 equals negative one. If we substitute these into the equation for circle 𝐴, we’re gonna have two minus nine all squared plus negative one minus three all squared, which is gonna give negative seven all squared plus negative four squared, which is gonna be equal to 49 plus 16, which is gonna give 65. Well, 65 is not equal to 100, so it’s not equal to the right-hand side of the equation. So therefore, we can rule out point (C).

So now, let’s move on to point (D). Well, for point (D), 𝑥 is equal to zero and 𝑦 is equal to two. So we’re gonna have zero minus nine all squared plus two minus three all squared. And that’s if we substitute these values into the equation for circle 𝐴 on the left-hand side. So this is gonna be equal to negative nine squared plus negative one squared, which is gonna be equal to 81 plus one, which gives us 82. And 82 is not equal to 100. So it’s not equal to the right-hand side of the equation. So therefore, we can say that this point does not lie on circle 𝐴, so, therefore, cannot be a point of intersection.

So finally, we could take a look at point (E). Well, point (E) has 𝑥-coordinate four and 𝑦-coordinate 12. If we substitute these into the equation for circle 𝐴, we get four minus nine squared plus 12 minus three squared. Well, this is gonna give negative five squared plus nine squared, which would give us 25 plus 81, which gives a final answer of 106. Well, 106 is not equal to 100, so once again not equal to the right-hand side of the equation for the circle 𝐴. So therefore, this again is not a point on the circle 𝐴.

So therefore, we can confirm that the point of intersection between circles 𝐴 and 𝐵 is the point (B) three, 11.