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5 points) Find the standard form for the equation of a circle h) + (y - k)? = 72 with a diameter that has endpoints of ( ~4,3) ad (9,5)_...

Question

5 points) Find the standard form for the equation of a circle h) + (y - k)? = 72 with a diameter that has endpoints of ( ~4,3) ad (9,5)_

5 points) Find the standard form for the equation of a circle h) + (y - k)? = 72 with a diameter that has endpoints of ( ~4,3) ad (9,5)_



Answers

In Problems 63-68, find the standard form of the equation of the circle that has a diameter with the given endpoints. $$ (-8,9),(12,15) $$

Okay, in this case was simply need to find the equation off circle. We have been given that in point off diameter in this open is given one comma minus seven. Right, Andi nine comma minus five. So four thing is very clear that the center co ordinate with the verge of this court. Next, events of one bless nine by two on minus seven minus fight back. Because that central light the midpoint off this in part of the diamond does have sent the coordinators 10 by two, which is five on minus tool. But do it is minus six, right? The two points are I'm just as human as one in minus seven on nine and minus five as the end point of the diamond. Right now the circle passes through this point B Q and we know that the question of circle is X minus its hold square less by minus. Kehoe it square. Is it wise to our square? So this buoyant will always decide this fire. The equation of the circle sent to this five and minus 61 minus five squid If I just can't see the point B White. So why in this good becomes minus seven and keys minus six. So this becomes plastics will square because minus minus is plus, that is a close to our square. So this simply comes out with 16 plus, um, minus seven, which is one that is eclipsed, Wallace. But so from here, I get them. I live are square. Right. So are square is equivalent to 70. No, the question of circle finally will be here. We have center, which is a five in minus X rays. So eggs minus five squared, plus by plus six square. That is a close to 70. All right.

We need to find the standard form of the equation of the circle that has a diameter with the given endpoints. So we first need to find our center so that's our midpoint. So we do that by adding or two X values divided by two, Adding R two y values And divided by two. So we label our points X one Y one X two Y 2. So we get five plus five divided by two. We get negative one plus seven divided by two. And this gives us five And this gives us three. So this is our center. So now we're going to find the distance from the center to one of the points because I would give us our radius. So we're going to call our center 53 and then we're going to use the other point 57 So this is going to be X one Y one X two, Y 2. So our distance formula which is our radius is equal to the square root of x two minus x one squared plus Y tu minus why one squared. So we have distance which is our radius parentheses, parentheses minus sine squared plus parentheses, parentheses minus sine squared. So we play in our values So X two is 5. X one is 5, why two is seven, Why one is three? So this simplifies to be The square root of zero squared plus four squared, Which ends up being the square root of 16, which is four. And again this is our radius. So our equation of our circle is x minus H squared plus y minus k, squared is equal to r squared, our center is our H and our K. So we have x minus five squared plus y minus three squared is equal to r squared, R squared wire is four, so r squared is 16.

We need to find the standard form of the equation of the circle that has a diameter with the given endpoints. So these are in points. We need to find our center so that's finding our midpoint. So the formula is X one plus X two divided by two. And we add our y. Values and divide by two. So we have X one Y one X two, Y 2. So this is negative six plus zero, divided by two and zero plus negative eight divided by two. So this simplifies to be -3 and -4. So that is our center. That's where center is at -3 -4. And we're going to use one of the other points So we'll use 0 -8. So this one's gonna be x one y 1 X- two, Y 2. So now we're going to find the distance from the center to the edge of the circle. So that's gonna be our radius. So the square root of X two minus X one squared plus y tu minus y one squared. Do we have parentheses? So we plug in our values X two is zero. X one is negative three. Y two is negative eight. Why one is negative four. So we simplify so our distance which is our radius is equal to three squared plus, negative eight minus and negative four groups negative eight minus negative four becomes negative four squared. So this becomes the square root of 9-plus 16 Which is the square root of 25 which is five. And this is again our radius. So now we're gonna find our equation of our circle. Sorry, equation of our circle is x minus H squared plus. Why minus K squared is equal to r squared? Our H. And our K. Come from our center and our our comes from our radius. So we have x minus a negative three squared plus why minus a negative four squared is equal to five square. So we clean this up because we can have double negatives. So our equation of our circle is X plus three squared plus Y plus four squared is equal to 25.

We need to find the standard form of the equation of the circle that has the diameter with the given endpoints. So we're going to label are points X one Y one X two Y 2. So we're going to find our midpoint which is our center. So that's adding our X. Values together, divided by two adding ry values together and dividing by two. So we have negative four plus six divided by two. And we have three plus three divided by two. So that gives us two divided by two which is one and then we get six divided by two which is three. So our center of our circle is that 13 So this is gonna be our H. And our K. And so we need to find the distance. So we're going to use the .63. So that's from the center to the edge of the circle. So this is gonna be our X one Y one. This is our X two Y two. R distance formula which this is going to be, our radius is equal to the square root of X two minus X one squared plus y tu minus y one squared. So we plug in our values so X two is six. X one is one. Y two is three why one is three? So we simplify, can we get the square root of five squared plus zero squared? So we get the square root of 25 which is five. So again this is our radius. So our standard form of our circle is x minus h squared plus y minus k squared is equal to r squared. So our H and our KR one and 3. So we're going to have x minus one squared plus y minus three squared is equal to r squared, R squared is five Squared, which is 25.


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