Explanation for Question 27 From the Math (Calc) Section on the Official Sat Practice Test 6
Question 27 asks us and the X, 2 Y plane, the graph of two X squared minus six X plus two Y 3 squared. Plus two, Y is equal to 45 is a circle. 4 What is the radius of a circle? Okay. 5 So to solve this problem, we need to know the general equation for a 6 circle in the X, Y plane. And that's going to be this expression right 7 here. We have X minus H squared, 8 plus Y minus K squared is equal to R squared. 9 What the center of the circle being H coma K, 10 and the radius of the circle, being that our term right here. 11 So I've copied down the expression that we're given right beneath it. 12 And we can see that we're pretty close to the same type of expression, 13 but we want to make ours look exactly like this so that we can 14 find out the value of our, it's not enough that we can't say that 15 ours 45, because we don't have the X's and Y's factor yet. 16 So looking at this, the first thing I'm going to want to do is 17 I want the coefficients in front of X squared and Y squared to be 18 one. So I'm going to divide both sides of the expression by two. 19 And that gives me X squared, minus three X plus Y 20 squared plus two, or excuse me, 21 Y squared plus Y is equal to 45 over two, 22 which is 22.5. So we're still not quite 23 there yet. And you might wonder, well, how can I factor this? 24 Because we don't have a constant on this side, 25 the way we're going to do that is we're going to complete the square. 26 So if you remember, you complete the square, where if you have ...