Please help me with my math Radius Homework Help is really appreciated!
PS: Pls Don't be bothered by my erasure on my sheet, those 4 question are unanswered.
Instructors' formula is x^2 + y^2 = r^2
$$\begin{array}{rcl} x^2 + y^2 &=& 49\\ x^2 + y^2-225 &=&0\\ 3x^2+3y^2 &=& 180\\ \frac{1}{5} x^2 + \frac{1}{5} y^2 &=& 25 \end{array} $$
It's always the same technique: all your equations have the same form with $a,c > 0,$
$$ax^2+ay^2 = c \iff x^2+y^2 = \frac{c}{a} \ (=r^2)\iff r = \sqrt{\frac{c}{a}} $$
Example: $8x^2+8y^2 = 32 \iff x^2+y^2 = 4 \iff x^2+y^2=2^2 \iff r = 2 = \sqrt{\frac{32}{8}}$
EDIT:Detailed explanation on how to find the center and the radius of a general circle. The general formula for a circle in the plane of radius $r>0$ and center $(c_1,c_2)\in \mathbb{R}^2$ is $$(x-c_1)^2+(y-c_2)^2=r^2.\qquad (*)$$ In you exercise all the circles are centered in $(0,0)$ so the expression reduces to $$x^2+y^2 = r^2.$$ Here's a method for going from a reduced circle equation to the canonical form $(*)$. I will explain it through an example, because the most difficulty is finding the parameters. So let us consider the equation $$ 2x^2+2y^2-4x+12y-30=0, \qquad (1)$$ we want to rewrite it in the form of equation $(*)$. Remember that $$(a-b)^2 =a^2-2ab+b^2 \qquad (**)$$, thus we want the in front of $x^2$ and $y^2$ to be $1$, therefore let us divide equation $(1)$ by $2$, this leads to
$$ x^2-2x+y^2+6y-15=0. \qquad (2)$$ Note that if the factors in front of $x^2$ and $y^2$ are not the same, then the set of $(x,y)$ satisfying the equation is not a circle. Still by identification with equation $(**)$ we now want to know what are the terms of the form $-2ab$, i.e. if $a=x$ or $a = y$ then what is $b$ such that we find $-2ab$ in equation $(2)$. Since the only terms of degree one in $(2)$ are $-2x$ and $6y$ it is easy to see that if $a=x$, then $-2x = -2ab$ for $b=1$ and if $a=y$, then $6y=-2ab$ for $b = -3$. So, observe that $$(x-1)^2 = x^2-2x+1=(x^2+2x)+1 \iff x^2+2x=(x-1)^2-1$$ and $$(y-(-3))^2=y^2+6x+9=(y^2+6y)+9 \iff y^2+6y=(y+3)^2-9,$$ Finally, substituting these relations in equation $(2)$ shows $$(x-1)^2-1+(y+3)^2-9-15=0 \iff (x-1)^2+(y+3)^2=25.$$ So we showed that $$2x^2+2y^2-4x+12y-30=0 \iff (x-1)^2+(y-(-3))^2=5^2,$$ and by identification with $(*)$ it follows that this is a circle with center $(c_1,c_2)=(1,-3)$ and radius $r = 5$.