Hi I need help with the following question
I have snapshotted the question at attached it to this thread.
any help is appreichated
thank you
Hello, rpatel!
Here's the first part.
I'll try to get back to the rest later . . .
There are a number of approaches to this problem.(a) Find the equation of the circle that passes through A(2,3), B(0,-1), C(-4,-3).
What is the center and radius of this circle?
I think this one is the simplest. (I may be the only one!)
The center is the point $\displaystyle P(x,y)$ equidistant from $\displaystyle A,B,C.$
Equate the distances: .$\displaystyle PA \:=\:PB \:=\:PC$
. . $\displaystyle \underbrace{\sqrt{(x-2)^2+(y-3)^2}}_{{\color{blue}[1]}} \:=\:\underbrace{\sqrt{x^2+(y+1)^2}}_{{\color{blue }[2]}} \:=\:\underbrace{\sqrt{(x+4)^2 + (y+3)^2}}_{{\color{blue}[3]}}$
From $\displaystyle {\color{blue}[1] = [2]}$, we have: .$\displaystyle x^2-4x+4 + y^2-6y+9 \:=\:x^2+y^2+2y+1 \quad\Rightarrow\quad x + 2y \:=\:3$
From $\displaystyle {\color{blue}[2] = [3]}$, we have: .$\displaystyle x^2+y^2+2y+1 \:=\:x^2+8x+16 + y^2+6y+9 \quad\Rightarrow\quad 2x + y \:=\:\text{-}6$
Solve the system of equations: .$\displaystyle x \:=\:\text{-}5,\;y \:=\:4$
. . Hence, the center is: .$\displaystyle P(-5,4)$
The radius is the distance $\displaystyle PA$ ... (or $\displaystyle PB$ or $\displaystyle PC).$
. . $\displaystyle r \;=\;PA \:=\:\sqrt{(\text{-}5-2)^2 + (4-3)^2} \:=\: \sqrt{49+1} \:=\:\sqrt{50} \:=\:5\sqrt{2}$
Therefore, the equation of the circle is:
. . $\displaystyle \left(x-[\text{-}5]\right)^2 + (y - 4)^2 \:=\:(5\sqrt{2})^2 \quad\Rightarrow\quad\boxed{ (x+5)^2 + (y-4)^2 \:=\:50}$
to c)
There are 6 parallelograms:
Take 2 points to determine one side of the parallelogram. Then you'll get the 4th vertex if you add to the staionary vector pointing to the 3rd vertex the vector which is defined by the 1st and 2nd vertex:
The side is CB. Then you get the coordinates of D:
$\displaystyle \overrightarrow{OA}+\overrightarrow{CB}=\overright arrow{OD_1}$
or
$\displaystyle \overrightarrow{OA}-\overrightarrow{CB}=\overrightarrow{OD_2}$
You can repeat this procedure with two more sides.
Hello again, rpatel!
I won't try to draw a diagram for this. .I hope you can visualize my reasoning.(b) Find the equation of the circle with center (10,10)
that just touches the circle you found in part (a).
The first circle is: .$\displaystyle (x+5)^2 + (y-4)^2 \:=\:50$
Its center is: $\displaystyle P(\text{-}5,4)$ and its radius is: $\displaystyle 5\sqrt{2}.$
The second circle has center $\displaystyle Q(10,10)$ and we must find its radius.
The distance $\displaystyle PQ$ is: .$\displaystyle \sqrt{(10-[\text{-}5])^2 + (10-4)^2} \:=\:\sqrt{225+36} \:=\:\sqrt{261} \:=\:3\sqrt{29}$
. . Hence, the radius of the second circle is: .$\displaystyle 3\sqrt{29}-5\sqrt{2}$
The equation is: .$\displaystyle (x-10)^2 + (y-10^2 \:=\:(3\sqrt{29} -5\sqrt{2})^2$