Use cramers rule to find the equation for a circle that...

Name: use cramers rule to find the equation for a circle that passes through the following points 4 3 2 5 5 2 69808
Uploaded: 2023-06-17T23:46:44-08:00
Duration: 16 min 27 s
Channel: Oswaldo Jiménez
Description: use cramers rule to find the equation for a circle that passes through the following points 4 3 2 5 5 2 69808

Transcript

00:01 We want to find the equation of the circle that passes through the points 95, 1, 93, 2 and 91, 1 more precisely what we're talking about here is the equation of the circumference passing through these three points and if you calculate the slope of the lines containing any two of these points you can see that they are different that is a slope from this of the line through these two points is different from the slope of the line through these two points i said that because it's easy to verify that these three points are not are not aligned that is they are not all three on the same straight line and for that reason we are sure we can find the equation of the circumference passing through the three points so let's say the center of this circle or the circumference we're going to call it we're going to call it point ab in the plane of the circumference let's say ab and we have the radius of the circle or the circle or circumference let's say radius okay radius r which is of course a positive number and so we know the equation of the circumference with these properties with this center and this radius will be x minus a square plus y minus b square equal r square that is x the variable x minus the first component of the center that difference squared plus y the second variable minus the second component of the center that is y minus b and that square and that sum of squares got to be equal to r square that comes from the definition of the circumference and it's a well -known equation so having this formula here which we want to determine this problem you can see that we have three values to calculate we have a to be calculated b and r and we will have three equations because each point here is on the circumference that is each point here verifies this equation here so negative five two sorry negative five one negative three two and negative one verify the equation of the circumference because all of them are on the circumference so we have a system of equations three equations which is the first component of the point negative five one that is negative five minus eight all that square plus the second component of the point negative five one that is one minus b square is r square same thing for the second point here is negative three minus a square plus two minus b square is equal to r square and for the third third point negative one minus a square plus one minus b square equal r square now we do some simplifications here and the first we see here is that we can get common factor out negative and because it's raised to the second power then this becomes a plus five square or five plus a it's the same plus one minus b square equal r square second equation we do the same we have a negative common factor out which is squared and so we get a plus three square plus two minus b square equal r square and the third equation the same thing here negative common factor out and we get a plus one square plus one minus b square equal r square and now we take this first and third equations because we see we have the same same part one minus b square here and here and on the right r square so we can say that the first equation minus the third equation give us a new equation which is we see that the term y minus one minus b square cancel out with one minus b square here and on the right r square minus r square cancel out giving us zero on the right so we get a plus five square here minus this term here minus a plus one square is all the all we get on the left of the equation of the difference of the first and third equation because the term one minus b square is repeated so it cancels out and that is equal to r square minus r square is zero so we have an equation of only a we can calculate a from here and for that we need to calculate both squares and take care that this negative sign is affecting all what we get from this square right here so we get a square plus 10a plus 25 that is this first square minus parentheses what we get in second square is a square plus 2a plus one…

Question

Use Cramer's rule to find the equation for a circle that passes through the following points: (4, 3) (-2, -5) (5, 2)

Question

Use Cramer's rule to find the equation for a circle that passes through the following points: (4, 3) (-2, -5) (5, 2)

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