This video demonstrates three methods to find the radius of a circle given a chord of length 4 that is bisected by a perpendicular line of length 1. Method 1 uses coordinate geometry by setting up the circle equation (x-H)² + (y-K)² = R² and substituting points M(0,0), N(4,0), and P(2,1) to solve for the center coordinates and radius. Method 2 applies the intersecting chords theorem, where the product of the segments of one chord equals the product of the segments of the other chord (2×2 = 1×(2R-1)). Method 3 uses the Pythagorean theorem in the right triangle formed by connecting the center to the chord's midpoint and endpoint. All three methods yield the same result: the radius is 5/2 or 2.5 units.
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Given this figure, we're required to find the radius of the circle. So, we have this chord where the chord length is four and it's being separated by this line which is one and this line is perpendicular to this line.
And also this part of the chord is equal of this line from here to here is equals to from here to here. So, with that, what is the radius of this circle?
So, I'm going to start by labeling the points.
Let's call this M.
Let's call this N and this let's call it P and this one Q.
Okay?
Let the center of the circle be O.
Now, since MN is four and they have told us this M MQ is equals to QN, that implies if this is four then MQ is two and QN is two.
Now, what you're going to use is what we call coordinate geometry to solve for this, okay? I'm going to draw the X and Y axis about point M. So, I'm going to draw a vertical line about point M and a horizontal line.
So, I've drawn the X axis.
So, this is the X axis and this one here is the Y axis.
Okay?
So, this point M is the origin, so its coordinates are zero zero since it's the origin. Now, since MN is four, so we consider that the X value at this point is 4 and Y is 0.
That is at point N.
Then at point P since from here to here is half of 4, so from here to here is we consider it the X value to be 2 and the Y value is 1. Okay?
So now at point M the coordinates are 0 0.
At point N the coordinates are 4 0. And at point P the coordinates are 2 1.
Now we know that equation of a circle the general equation is X - H squared plus Y - K squared is equals to R squared where H and K is the center.
And R is the radius of the circle.
So at this center here that's where we have the coordinates H which uh H belongs to X axis and K.
Okay?
So now for point M which is 0 0.
If we are to replace in our equation of the circle which is X minus H squared plus Y minus K squared is equals to R squared, our X this is x and this is y.
So, x value is 0, so 0 - x squared plus also y is 0, so 0 - k squared is equals to r squared.
This gives us h squared plus k squared being equals to r squared. Let's call this equation one.
Now, for point n n was 4 0.
>> [snorts] >> Replacing in our equation which is x minus h squared plus y minus k squared is equals to r squared. This is our x and this is our y.
So, we shall have 4 minus h squared plus 0 minus k squared being equals to r squared.
We know that from a minus b squared being equals to a squared minus 2ab plus b squared.
So, this one here becomes 16 minus 8h plus h squared plus this one here becomes k squared is equals to r squared. So, we rearrange, we shall have h squared plus k squared minus 8h plus 16 is equals to r squared. Let this be equation two.
Then for point p which is 2 1.
This is our x and this is our y.
Replacing in our equation of the circle x minus h squared plus y minus k squared is equals to r squared x is 2 minus h squared plus 1 minus k squared is equals to r squared. From the other formula that we saw of a minus b squared is equals to a squared minus 2 a b plus b squared, so this becomes 4 minus 4 h plus h squared then plus this becomes 1 minus 2 k plus k squared is equals to r squared. When we rearrange, we shall have h squared plus k squared minus 4 h minus 2 k So, we have 4 plus 1 which is plus 5 is equals to r squared. Let this be equation three.
The next step is let's substitute equation one into equation two.
Okay? So, our equation one was h squared plus k squared is equals to r squared and our equation two is h squared plus k squared minus 8 h plus 16 is equals to r squared. So, according to equation one, h squared plus k squared is r squared, so this here becomes r squared. So, r squared minus 8 h plus 16 is equals to r squared. Of course, when this comes crosses this side, they cancel.
So, we're left with minus eight h plus 16 is equals to zero.
So, we shall have negative eight h being equals to negative 16 divided by negative eight.
We shall have h being equals to positive two.
Okay?
So, now let's substitute equation equation one into equation three.
So, our equation three was h squared plus k squared minus four h minus two k plus five is equals to r squared and we said h squared, according to equation one, h squared plus k squared is equals to r squared.
So, this here is r squared minus four h minus two k plus five is equals to r squared. This cancels with this.
So, we are left with minus four h minus two k plus five is equals to zero. Take this five this side, we shall have minus four h minus two k being equals to negative five.
But, from here we've got h to be equals to two.
Substitute it here, we shall have negative four times two minus two k is equals to negative five. This gives us negative eight minus two k is equals to negative five.
So, we shall have negative two k being equals to minus five plus eight.
So, negative two k is equals to This gives us three.
Okay, divide by -2.
-2 and we are left with k to be equals to negative 3 over 2.
So, we have got our value of h as 2 and our value of k as -3 over 2.
Okay?
Now, from equation one where h squared plus k squared is equals to r squared, we have h and we have k. So, now I can find r, where r is the radius, okay? Which we is required.
So, this will be 2 squared plus this is -3 over 2 squared is equals to r squared.
So, 2 squared is 4 and plus -3 squared is 9 over 2 squared is 4.
This is equals to r squared. Here we find the LCM and the LCM is 4.
4 by 1 is 4. 4 * 4 is 16 plus 4 by 4 is 1. 1 * 9 is 9. is equals to r squared.
16 plus 9 is 25. So, 25 over 4 is equals to r squared.
Therefore, r squared is 25 over 4. We find the square root on both sides.
Therefore, r square root of 25 is 5 and square root of 4 is 2. So, r is 5 over 2 or we can say r is 2.5 units.
So, that's one way of finding the radius by using coordinate coordinate geometry.
Now, let's look at another method which we could have used to find still the radius of the circle.
So, in method two, let us draw first of all a line from point Q through the center this center O through the center O all the way to the bottom so that we come up with a diameter.
Okay?
Just like this. Okay? So, let me call this point S. Remember, this is four, so this is two and this is two.
So, if the radius of this circle is R, then that implies that PS which is the diameter will be 2R.
Then, QS will become since PQ is one, so QS will become 2R - 1.
Now, we know that from intersecting chords theorem, if you have this, then AB is equals to CD.
D.
So, in our diagram, our A is MQ which is equals to two.
Our B is QN which is equals to two as well.
And our C is PQ which is equals to one. And our D is QS which is equals to 2R 1.
Therefore, substituting into the formula AB is equals to CD, our A is two times our B is two is equals to one times two R minus one.
So, two times two is four is equals to two R minus one.
So, two R would be equals to four plus one.
Therefore, two R is equals to five.
Divide both sides by two.
And so, we get R to be equals to five over two, which is the same as 2.
five units.
So, that's another way in which we could find the radius of this circle.
Now, let's look at the third method in which we could have used to solve for the radius of this circle.
So, first we're going to connect uh center O to point Q and also connect this center to point N.
Just like this. So, this is 90° of course.
Now, as you can see, ON is the radius of this circle, which we are going to call R.
So, if ON is the radius, also OP is also the radius of the circle. So, ON is also the radius. Now, if OP is So, if OP is the radius, then that implies that and PQ is one, this implies that QO is equals to radius R minus what we have, one.
Okay? So, from here to here is R minus one.
Therefore, focusing on triangle QNO.
So, this is QN.
O, this is 90°. This is r minus one.
This is two, and this is r. Since it's a right triangle, we can use Pythagoras theorem to solve for r.
So, a squared plus b squared is equals to c squared.
Okay?
So, our a is two.
So, two squared plus our b is r minus one.
So, squared is equals to our c, which is r. So, r squared. Two squared is four, but we know that from a minus b squared is equals to a squared minus two ab plus b squared.
Therefore, two squared is four plus this here in this form becomes r squared minus two r plus one.
This is equals to r squared. Now, when this r squared crosses, it cancels with this.
So, we are left with four four minus two r plus one is equals to zero.
So, we have negative two r.
Four plus one is five. Plus five is equals to zero. So, we have negative two r equals to when this crosses, it becomes negative five.
So, we divide both sides by -2.
And therefore, r is equals to 5 over 2.
Or r is equals to 2.5 units.
So, that's how you solve for the radius of this circle. You can use either of the three methods to solve, whichever is simpler for you. But, it's important to try and know all possible methods, just in case.
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>> [clears throat] >> Bye.
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