The general equation for a circle,

, falls out of the Pythagorean Theorem, which states that the square of the hypotenuse of a right triangle is always equal to the sum of the squares of its legs (you might have seen this fact written like

, where <em>a </em>and <em>b</em> are the legs of a right triangle and <em>c </em>is its hypotenuse. When we fix <em /><em>c</em> in place and let <em>a </em>and <em>b </em>vary (in a sense, at least; their values are still dependent on <em>c</em>), the shape swept out by all of those possible triangles is a circle - a shape defined by having all of its points equidistant from some center.
How do we modify this equation to shift the circle and change its radius, then? Well, if we want to change the radius, we simply have to change the hypotenuse of the triangle that's sweeping out the circle in the first place. The default for a circle is 1, but we're looking for a radius of 6, so our equation, in line with Pythagorus's, would look like

, or

.
Shifting the center of the circle is a bit of a longer story, but - at first counterintuitively - you can move a circle's center to the point (a,b) by altering the x and y portions of the equation to read:
Answer:
Step-by-step explanation:
2(6²)-3(6)+2=y
(2x36)-18+2=y
72-18+2=y
y=52 BODMAS
It may also be 56 tho im not sure if thats right
Answer: nope
Step-by-step explanation: using the given coordinate, that would make the problem become 4 = 2(6) which then becomes 4 = 12
So, no, it does not make the equation true.
Hope this helps!
Answer:

Step-by-step explanation:
If G is the midpoint of CD, and AC is parallel to DB, then AC = DH.
Therefore, G is the midpoint of AH and ΔACE is similar to ΔDBE.
As AC : DB = 1 : 3
⇒ Area of ΔACE : Area of ΔDBE = 1² : 3² = 1 : 9
We are told that Area ΔACE = Area ΔAEG.
⇒ Area ΔACG = 2 × Area ΔACE
As AC = DH, and G is the midpoint of CD:
⇒ ΔACG ≅ ΔHDG
⇒ Area ΔHDG = 2 × Area ΔACE
Area of quadrilateral EGHB = Area of ΔDBE - Area ΔHDG
= Area of ΔDBE - 2 × Area of ΔACE
Therefore:


Using the ratio of Area ΔACE : Area ΔDBE = 1 : 9

