00:01
We have two parts in this exercise.
00:04
Part 1, consider the circles with the given equations.
00:08
For each circle, we want to find the coordinates of the center, the radius, the coordinates of the endpoints of the vertical and horizontal diameter.
00:20
Circles are in part a, x plus 1 square plus y minus 3 square equal 49.
00:25
In par b x square plus y plus five square equal 121 and in part c x minus 7 square plus y square equal 81 then in part two want to find the equation of the circle that has diameter with endpoints a 82 and b negative 6 2 so first before we solve the part 1 we uh we uh are going to write here some generalities about circle.
01:03
So first the general equation is given as x minus x0 square plus y minus y 0 square equal r square.
01:18
So in this formula here we have the center of the circle which is the point x0 y0 and the radius is r.
01:35
And as a you can see there's not a function because we we have points with two images on the curve that's why we get in fact a formula where we if we try to put y in terms of x we have two parts in fact two functions the upper part and lower part of the circle that's for the equation the center and the radius and now we have considering for example we have a circle like this and then there is the coordinate axis somewhere, such a way that we have the center, x0 -y -0, like this.
02:26
Then you have a diameter which is parallel to the x -axis.
02:33
There's only one with that property, for example, this, in this sketch.
02:40
And so there are, for that segment, which is parallel to the x -axis, we have two points on the circumference and those are the endpoints of the horizontal diameter and we have a similar diameter which is now parallel to the y -axis this one here and the endpoints of this segment that are on the circumference are the endpoints of the vertical diameter so that's it so we want to find those four points even any circle so let's start with first circle part a which is given us x plus one square plus y minus three squared equal 49 and we can rewrite this formula as x minus negative 1 squared plus y minus 3 square equal 7 square and written this way we can compare it to general formula here and we see immediately that the center of the circle is negative 1 3 and the radius is 7 so that's it for that part and now we can find the endpoints of the vertical and horizontal diameters.
04:44
So the particularity, for example, in the horizontal diameter like this, in the blue segment, is that the y value of each point on that segment is constant, the same value, and is just the second coordinate of the center of the circle.
05:07
And for the vertical diameter, we have that the value that is constant now for those points on that segment is x0, that is the first coordinate of the center.
05:22
So to find the end points of the horizontal diameter, let's say first.
05:40
So all we all we got to do is in the equation of the circle put, we are talking on the horizontal diameter.
05:48
With this one here.
05:51
So we put y0 is the value of all of all the second coordinates of the point.
05:58
So y is equal to y0.
06:02
Y is equal in this case to three, which is the second coordinate of the center.
06:10
And that implies in the general equation of this of this circle that x plus y square plus 3 minus 3 square equal 49.
06:24
That is, we have put the value of the constant value y of all the points on the horizontal diameter.
06:33
And that is this equation, which simplified because this is equal to 0, to x plus 1 square equal 49.
06:45
And if we take square root both sides of this equation, we get this.
06:51
And now you get to be very careful because the square root of x plus one square is the absolute value of x plus one.
07:02
That is because the quantity inside parentheses can be even negative, but the square is always positive.
07:10
So we have to consider the absolute value.
07:15
It's very different when we have something like this, because in this equation, x got to be from the start a positive value.
07:24
Cannot be negative.
07:25
So this is equal to x and there's no other possibility.
07:31
In this case, if you take a negative value of x, we can still get the square root of that, the square of that value be a solution.
07:42
Okay, so we get then this after value of x plus 1 equal to 7 and so from here we get two options.
07:52
That is we can have x plus 1 is directly equal to 7 or x plus 1.
07:57
Plus 1 is negative 7.
08:00
In each of these cases, x plus 1 absolute value will be 7.
08:06
From the first equation we get x equals 7 minus 1, that is 6, and in the second case, x equal negative 8.
08:17
So, because we know the second coordinates of any points of the horizontal diameter is 3, we can say that the 2 end points have the same second coordinate 3, and the first coordinate that 6 are 6 and negative 8 these endpoints of course i'm referring to this endpoints to the endpoints of the horizontal diameter are 6 3 and negative 8 3 and now the endpoints of the vertical diameter and those points are obtained by considering that in the vertical, let me get rid of this for a moment, the vertical diameter that is this one here, we have the same x value for all the points on that segment and that value is x0.
09:51
So in this case is x equal to negative 1, which is the first coordinate of the center, so it's negative 1, x equal negative 1.
10:03
And if we put that on the equation of the circle we get negative 1 plus 1 square plus y minus 3 square equal 49 and so because this is 0 we get y minus 3 square equal 49 and so the absolute so the square root of y minus 3 square equals square root of 49 so also the value of y minus 3 equals 7 and from this we have the two options.
10:43
Y minus 3 equals 7 or y minus 3 equal negative 7.
10:48
From the first equation we get y equal 10.
10:56
From the second, y equal 3 minus 7 is negative 4.
11:01
And all the points have the same first coordinate negative 1.
11:06
And so do the endpoints of the vertical diameter.
11:10
So they are negative 1, first coordinate here.
11:27
10 and negative 1 again, negative 4.
11:35
Here, remember, it's negative 4, clearer.
11:42
So these are the end points of each of the diameter.
11:47
Now with that, go to the second, sorry, not 2, but b, second circle.
11:57
Second circle has equation x square plus 1 y plus 5 square equal 121 and we write it like this x is x minus 0 square plus y plus 5 is y minus negative 5.
12:27
Remember i'm doing this and sort of before and let me in this part.
12:34
101 is 100, sorry it's 11 square.
12:38
Now i'm doing these transformations in order to have always the variable minus a value because that's the general formula of the circle.
12:47
The variable x, for example, minus a value and that square and the value of the variable y minus a value in that square and then equal to a value square.
12:59
So the value that is subtracting from the x is the first coordinate of the center and the value that is substracting.
13:08
The y is the second coordinate of the center and the value that is squared on the hand side is the radius so that's why i'm always transformed made in this given equations to meet directly that form and so when we do that we can see directly the center and the radius in this case the center is zero which is a value that is subtracting x in the term where the variable is is x and the value that is subtracting to y is negative 5 and that's the center the radius is the value that is square on the right is 11 so we have that now we calculate the end point of the horizontal horizontal diameter is better horizontal diameter okay and we know in the horizontal diameter, the y is constant, so it's constant equal to the second coordinate of the center, negative 5.
14:47
We put that into the equation of the circle and we get x square equal 121 directly because we know that that value of y is going to nullify the second term of the equation, that is the term y plus 5 square.
15:04
And from here we get the square root of x square equal square root of 121.
15:10
And we know that the square root of the square of a number is the absolute value of the number.
15:15
In this case, in this case, absolute value of x, and the square root of 121, because 121 is 11 square is 11.
15:25
And so we have two options.
15:29
X is 11 or x is negative 11.
15:33
And we have directly the first coordinates of the two end points of the horizontal diameter...
