hey guys this looks like a fun one we're given four squares inside of a circle and it wants to know what is the area of the circle if you want to try this on your own pause it right now cuz I'm going to solve it in 3 2 1 first I want to identify these points where the squares touch the circle and I also want to find the side lengths of these squares since the area is equal to 16 each of these sides will be equal to four next I want to look at some formulas for circles we have the area of a circle is equal to PK R 2 and we also have the equation of a circle is equal to X h^2 + y - k^2 = R 2 where h k is the center and the radius is equal to R so to answer the question we need to find the area of this circle next let's look at this other piece of notes this is the equation of a circle if we have a rectangular coordinate system and we want to graph a circle this is the equation to do it for example let's look at a specific Circle let's do this one right here that has a center at 65 and a radius of three in the place of the H we plug in the six in the place of the K we plug in five and in place of the r we plug in three this is the equation for this circle now let's take our Circle and stick it on the coordinate plane and I want to move the circle to align this intersection point with 0 0 next let's identify this point well the x coordinat is going to be equal to zero and the height of this point is the same as the square so this point is going to be at 04 and let's also find the coordinate for this point to find the x coordinate let's see how far back we go on the x-axis we have 4 8 12 and 16 so the x coordinate is going to be- 16 and then to get the y-coordinate we're going to go down one side length which is equal to 4 so this is going to be -4 now we have three points on our Circle let's find the equation of the circle that goes through those three points let's take this equation and let's copy it down three times for the first one we're using the point 0 0 so let's plug in 0 for the X and 0 for the Y for the point 04 let's plug in 0 for the X and four for the Y and then for this point in the place of the X let's plug in -16 and in the place of the Y let's plug in four now we have three equations and three variables let's solve for those variables all three of these zeros can disappear a negative * a negative is a positive so the quantity h^2 is the same thing as positive h^2 and the same thing for the k^2 and the negative h^2 they change into positive k^2 and positive h^2 next let's focus on these top two equations these are both equal to R 2 so we can set them equal to each other so we have h^2 + k^2 is equal to h^2 + 4 - k^2 for the first let's subtract h^2 from both sides of the equation we end up with k^2 is equal to the quantity 4 - k^ 2 4 - k^2 is equal to 4 - K * 4 - K and after we multiply that out we get 16 - 8 k + k^ 2 now we have a k^ 2 on both sides of the equation let's subtract k^2 from both sides of the equation on the left hand side we have zero and on the right hand side we have 16 - 8 K next let's add 8 K to both sides of the equation and we end up with 8 K is equal to 16 let's divide both sides by 8 and we have k is equal to 2 and now that we know that K is 2 we can change all of these KS into two let's clean things up a little bit let's start inside these parentheses 4 - 2 is = to 2 and -4 - 2 isal to -6 next we can take care of these squares 22 is equal to 4 2^2 is equal to 4 and -62 is equal to 36 these top two equations now match let's get rid of one of them and for these two equations these are both equal to R 2 so let's set them equal to each other on the left left hand side this -16 - h^ 2 is the same thing as -16 - H * -16 - H after we multiply this out it becomes h^2 + 32h + 256 and then we can bring down everything else we have an h^ squ on both sides of our equation so let's subtract h^2 from both sides of our equation and I'm thinking in the same step let's also subtract four from both sides on the left hand side let's bring down the 32h and 256 + 36 - 4 is equal to 288 and on the right hand side all this stuff cancels to give us zero let's subtract 288 from both sides of the equation that'll give us 32h is equal to - 288 and then after we divide both sides by 32 we get H is equal to9 now we can plug in9 for all the H's this x - H will become an x + 9 this H will become a 9 and this negative H will become a + 9 now let's clean things up again the quantity 9^ 2 is equal to 81 and 81 + 4 is equal to 85 this technically is all we need to finish but I kind of want to make sure that this equation will match inside the parentheses -16 + 9 is equal to -7 and -72 is equal to 49 49 + 36 is equal to 85 and both our equations give us the same thing that 85 is equal to R2 we can now update this r s to be 85 and this R squ to be 85 and Pi * 85 is the same thing as 85 Pi let's give it a label of square units and let's put a box around it how exciting