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.
Best Answer
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.