Question

What is the radius of the circle write your sneer in simplified, rationalized form

Answer 1

The equation of a circle is expressed as

`\begin{gathered} (x-a)^2+(y-b)^2=r^2\text{ -----equation 1} \\ where \\ (a,b)\text{ is the center of the circle} \\ r\text{ is the radius of the circle} \end{gathered}`

Expanding equation 1 gives:

`\begin{gathered} x^2+y^2-2ax-2by+(a^2+b^2)=r^2 \\ \Rightarrow x^2+y^2-2ax-2by+(a^2+b^2)-r^2=0\text{ ------- equation 2} \end{gathered}`

Given the equation of the circle to be expressed as

`x^2+y^2-y-12=0\text{ ----- equation 3}`

Comparing equations 2 and 3, we have

`\begin{gathered} \Rightarrow-2ax=0\text{ ---- equation 4} \\ \Rightarrow-2by=-y\text{ ----- equation 5} \\ \Rightarrow(a^2+b^2)-r^2=12\text{ ----- equation 6} \end{gathered}`

From equation 4,

`\begin{gathered} -2ax=0 \\ divide\text{ both sides by -2x} \\ \Rightarrow a=0 \end{gathered}`

From equation 5,

`\begin{gathered} -2by=-y \\ divide\text{ both sides by -2y} \\ -(2by)/(-2y)=-(y)/(-2y) \\ \Rightarrow b=(1)/(2) \end{gathered}`

From equation 6,

`\begin{gathered} (a^2+b^2)-r^2=-12 \\ where\text{ } \\ a=0,\text{ b=}(1)/(2) \\ \Rightarrow(0^2+((1)/(2))^2)-r^2=-12 \\ (1)/(4)-r^2=-12 \\ subtract\text{ }(1)/(4)\text{ from both sides of the equation,} \\ (1)/(4)-r^2-(1)/(4)=-12-(1)/(4) \\ \Rightarrow-r^2=-(49)/(4) \\ divide\text{ both sides by -1} \\ \Rightarrow r^2=(49)/(4) \\ take\text{ the square root of both sides,} \\ √(r^2)=\sqrt{(49)/(4)} \\ √(r* r)\text{ =}\sqrt{(7)/(2)*(7)/(2)} \\ \Rightarrow r=(7)/(2) \end{gathered}`

Hence, the radius of the circle is

`(7)/(2)`