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1) Find the equation of the circle of radius $2$ with center at $(3, 0)$.

My answer: $\sqrt{(x-3)^2 + y^2} = 2$

2) Find the equation of the circle of radius $\sqrt3$ with center at (-1, -2).

My answer: $\sqrt{(x+1)^2 + (y+2)^2} = \sqrt3$

3) There are two circles of radius 2 that have centers on the line $x = 1$ and pass through the origin. Find their equations.

My answers: $\sqrt {(x-1)^2 + (y+\sqrt3)^2 }= 2$, $\sqrt {(x-1)^2 + (y-\sqrt3)^2 }= 2$.

4) Find the equation of the circle that passes through three points , $(0, 0)$, $(0, 1)$, $(2, 0)$.

My answers: The three points $(0, 1)$, $(0, 0)$, $(2, 0)$ make a right-angled triangle at $(0, 0)$. According to the Thales' theorem, the hypotenuse of the triangle which is a line-segment from the point $(2, 0)$ to $(0, 1)$ is the diameter of the circle. The center "P" lies on the point $(0+2/2, 1+0/2)$ = $P(2, 1/2)$. Therefore, the circle will be the locus of the equation: $\sqrt {(x-1)^2 + (y-1/2)^2} = r$

5) Find the equation of the circle one of whose diameters is the line segment from $(-1, 0)$ to $(5, 8)$.

My answers: $\sqrt {(x-2)^2 + (y-4)^2} = 5$

If any of the answers is wrong, so tell me the correct one please.

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To make your answer readable, user curly braces for what you wants "inside" the square root symbol. For example, compare $\sqrt x+y$ which is \sqrt x+y with $\sqrt{x+y}$ which is \sqrt{x+y}. It is also nicer if you write the equations in the form $(x-k)^2+(y-h)^2=r^2$ where $(k,h)$ is the center and $r$ the radius, which avois surds altogether. – Pedro Tamaroff Mar 31 '13 at 4:00
@PeterTamaroff Thank you Peter for editing my answers, it would help one to understand my answers more easily. – Samama Fahim Mar 31 '13 at 4:04

First one is correct, the rest is not. For the second one, your mistake is minor, it is just $1.73^2 \neq 3$. For the third one, I suggest you draw the Cartesian plane and then see which two circles you are looking for. For the fourth one, you have listed the conditions that your equation should satisfy, but did not write the equation itself. Again, I suggest you proceed from the Cartesian plane. Again, for the fifth one, just draw it and the answer will be immediate.

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2) $\sqrt{(x+1)^2+(y+2)}=\sqrt 3$ (You should not replace $\sqrt 3$ by $1.73$)

3) $(S_1): (x-1)^2+(y-\sqrt 3)^2=4$ and $(S_2): (x-1)^2+(y+\sqrt 3)^2=4$

(Hint. The center is the intersecrion of the circle with origin center, radius=$2$ and the line $x=1$. There are two intersection points $I_1=(1,\sqrt 3)$ and $I_2=(1, -\sqrt 3)$)

4) $(S): (x-1)^2+(y-\frac{1}{2})^2=\frac 5 4$

(Hint. Three points make a rectangle triangle at (0, 0). Hence, Radius $I=(\frac{0+2}{2},\frac{1+0}{2})=(1, \frac 1 2)$, Radius $R=\frac{\sqrt{(0-2)^2+(1-0)^2}}{2}=\frac{\sqrt {5}}{2}$))

5) $(S): (x-2)^2+(y-4)^2= 25$

(Hint. Center $I=(\frac{-1+5}{2},\frac{0+8}{2})=(2, 4)$, Radius $R=\frac{d}{2}=\frac{\sqrt{(5-(-1))^2+(8-0)^2}}{2}=5$)

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how did you figure out for the fourth problem that (0,1) and (2, 0) constitute the diameter of the circle? – Samama Fahim Mar 31 '13 at 13:54
I think we have to assist our understanding with a theorem of circle which says: "If a right-angled triangle is inscribed in a circle, and if the sides adjacent to the right-angle are touching any two points on the circle, two points for each side, then the hypotenuse of the triangle will be the diameter of the circle." – Samama Fahim Mar 31 '13 at 15:20

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