In Exercises 65-66, a line segment through the center of each cir...

Question

In Exercises 65-66, a line segment through the center of each circle intersects the circle at the points shown. a. Find the coordinates of the circle's center. b. Find the radius of the circle. c. Use your answers from parts (a) and (b) to write the standard form of the circle's equation.

Accepted Answer

Hello today we're going to be identifying the radius and center of the given circle. And we're going to be using that information to write the standard form of the equation of the circle. So we can go ahead and identify the center by just looking at it graphically, we know that the center is located right in the middle of this equation of the line. So it's gonna be located right here. And in order to reach this point you're going to want to travel four units to the left on the X axis and you're going to want to travel up seven units on the y axis. So that means that the coordinate for the center of the circle is going to be negative for comma seven. So again the center for the circle is going to be negative for comma seven. And since we know that the center of the circle is not at the origin, the standard form of the equation of the circle is going to be x minus H squared plus y minus k squared. And that is going equal to r squared. And this is where the center is equal to h comma K. So just using the center means that H is going to equal to negative four and K is going equal to positive seven. So we have our H and K for the standard equation and now we just need to find the radius while the radius is only half the distance of the diameter and we have the location or the coordinate for one of the endpoints of the radius and we have the coordinate for the center of the circle. So we can take these two points and plug them into the distance formula. That way we can go ahead and find the distance of the radius. So just just to recall the distance formula is the square root of X two minus x one, quantity squared plus y two minus y one quantity squared. And the two points that we're going to be using is the center which is negative four comma seven and the end point negative three comma five. Here we can go ahead and let negative four comma seven, B r x one Y one and we can let negative three comma five br x two, Y two. So if we plug this into the distance formula what we get is the square root of x two minus x one which is negative three minus negative four which is plus four quantity squared plus y two, which is five minus y one which is seven quantity squared. And now we just need to go ahead and simplify here, negative three plus four is going to give us positive one and positive one squared is just going to give us 15 minus seven is going to give us negative two and negative two squared is going to give us positive four. So we have the square root of one plus four and one plus four is going to give us five which is going to give us the square root of five. Now for the square root of five cannot be reduced any further. So this is going to be the distance of the radius or in other words the radius is going to equal to the square root of five. But keep in mind that for the standard equation of the circle we want r squared. So we're gonna go ahead and square both sides of r is equal to the square root of five. And this is going to give us r squared is equal to five, which is the last piece of information that we need. So now we're gonna go ahead and plug this in to the standard form of the equation of a circle. That's going to give us x minus h which h is negative four squared plus y minus k. Which is going to be y minus seven squared equal to r squared which is five. And just to simplify x, the x quantity negative one times negative four is going to give us positive four. So what we have here is x plus four squared and this is going to be the standard form of the equation of the circle. So just to go ahead and summarize, the standard form of the equation of the circle is x plus four squared plus y minus seven squared equal to five. The center is at negative four comma seven and the radius is equal to the square root of five. With that being said, the answer to this problem is going to be C. So I hope this video helps you in understanding how to approach this problem, and I'll go ahead and see you all in the next video.

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In Exercises 65-66, a line segment through the center of each circle intersects the circle at the points shown. a. Find the coordinates of the circle's center. b. Find the radius of the circle. c. Use your answers from parts (a) and (b) to write the standard form of the circle's equation.