Skip to 0 minutes and 12 seconds Hello. Welcome to the section “It’s your turn on Distance and Circles”. Good. In this exercise we have to find the equation of the circle passing through the points A, B, and C. In particular, we want to compute the center and the radius of this circle. Good. Let us recall first of all which is the equation of a circle of center C, of coordinates x0, y0, and radius r. The equation is x minus x0 squared plus y minus y0 squared equal to r squared. We have to find the center and the radius such that this equation has the coordinates of the point A, B, and C as solutions. Then we consider the following system of equations. OK.

Skip to 1 minute and 21 seconds The coordinates of A, 2 and minus 1, has to be solutions of this equation. OK. Then we must have that 2 minus x0 squared plus minus 1 minus y0 squared has to be equal to r squared for suitable coordinates x0 and y0 of the center and radius r. The same for the point B. We have that minus 2 minus x0 squared plus 3 minus y0 squared has to be equal to r squared. And finally also for the point C. And we have 0 minus x0 squared plus 5 minus y0 squared equal to r squared. OK. Let us do some easy computations.

Skip to 2 minutes and 22 seconds Then the first equation is we have x0 squared minus 4 times x0 plus 4 plus y0 squared plus 1 plus 2 times y0 equal to r squared. OK. The second equation now. x0 squared plus 4 times x0 plus 4 plus y0 squared plus 9 minus 6 times y0 equal to r squared. And finally the third equation, which is x0 squared plus y0 squared minus 10 times y0 plus 25 equal to r squared. Good. Let us call first this equation, second this equation, and third the last equation. And now if we subtract from the first equation the second equation, what do we get? OK. We get the first equation minus the second equation is x0 squared minus x0 squared is 0.

Skip to 4 minutes and 6 seconds Here we have minus 8 times x0 then we have the constant term of 4 plus 1 minus 13, which is minus 8.

Skip to 4 minutes and 22 seconds And then we have this minus this is 0. This minus this is plus 8 times y0 equal to r squared minus r squared equal to 0. And now we consider the first equation minus the third equation. And we have this minus this, which is 0. And then we have minus 4 times x0. Then this minus this is 0. Then we have 2 y0 minus minus 10 y0 is plus 12 y0. And finally we have 4 plus 1, 5, minus 25, which is minus 20 equal to r squared minus r squared equals 0. And then we leave the third equation, which is x0 squared plus y0 squared minus 10 times y0 plus 25 equal to r squared. OK.

Skip to 5 minutes and 43 seconds Look, we have transformed our system of equations in a new system of equations. But this procedure is clearly reversible. Indeed, you see if we have this system of equations, then you easily get the first old equation how? OK. Just summing to this equation, which is 1 minus 3, the equation three. And you get the equation one. Then when you have the equation one, then you can obtain the equation two how? Subtracting from the equation one the equation one minus two. And the third equation is exactly the same as before. Because the procedure is reversible, then these two systems of equations are equivalent. They have the same solutions. OK.

Skip to 6 minutes and 40 seconds Now let us simplify a little this system and what do we get? OK. We can divide by 8 the first equation and we have minus x0 minus 1 plus y0 equal to 0. And we can divide by 4 this second equation. And we have minus x0 plus 3 times y0 minus 5 equal to 0. And again the third equation. x0 squared plus y0 squared minus 10 times y0 plus 25 equal to r squared. Good. And now let us subtract from the first equation the second equation. And we get this minus this is equal to 0. Then we have minus 1 minus minus 5 is like minus 1 plus 5, which is 4.

Skip to 7 minutes and 47 seconds And then y0 minus 3 times y0 is minus 2 times y0 equals 0. And this implies immediately what? That y0 has to be equal to 2. Good. And now consider the second equation substituting y0 equals 2 in this equation. And what do we get? Minus x0 plus 3 times y0 is plus 6 minus 5 equal to 0. And then we immediately get x0 equal to 1. Good? And finally we have now the third equation. And we can substitute here, here, and here to x0 and y0 the two values that we have found. And what do we have?

Skip to 8 minutes and 43 seconds x0 squared, which is 1 squared, which is 1, plus y0 squared, which is 4, minus 10 times y0, which is minus 20 plus 25 equal to r squared. And we get that r squared has to be equal to 25 minus 20 is 5 plus 4 plus 1 is 10. That is, r has to be equal to the square root of 10. Then we have found our circle is a circle of radius the square root of 10 and of center the point of coordinates 1, 2. The center is the point of coordinates 1, 2, the radius is the square root of 10, and then its equation is x minus 1 squared plus y minus 2 squared equal to 10. Good.

Skip to 9 minutes and 54 seconds Thank you very much for your attention.

# It's your turn on distance and circles

Do your best in trying to **solve the following problems**. In any case some of them are solved in the video and all of them are solved in the pdf file below.

### Exercise 1.

Find the equation of the circle passing through \(A=(2,-1)\), \(B=(-2,3)\), \(C=(0,5)\) and compute its center and its radius.

### Exercise 2.

For which values of the parameter \(k\) does the equation \(3x^2+3y^2-4x+6y-k=0\) represent a circle? For each of these values of the parameter, compute the center and the radius of the obtained circle.

### Exercise 3.

Find the equation of the circle of center \(C=(5,-1)\) and passing through \(A=(-2,3)\).

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