Question

*URGENT + 100 POINTS*

A right triangle is shown in the graph.

right triangle on coordinate plane with hypotenuse labeled t and one endpoint of hypotenuse at r comma s and the other endpoint at x comma y, vertical line from point x comma y and horizontal line from r comma s that meet at right angle of triangle, horizontal dotted line from point r comma s to point s on y axis, horizontal dotted line from point x comma y to point y on y axis, vertical dotted line from point r comma s to point r on x axis, and vertical dotted line from right angle to point x on x axis

Part A: Use the Pythagorean Theorem to derive the standard equation of the circle with center at (r, s) and a point on the circle at (x, y). Show all necessary math work. (3 points)

Part B: If (r, s) = (7, –4) and t = 10, determine the domain and range of the circle. (4 points)

Part C: Is the point (9, 1) inside the border of the circle if (r, s) = (7, –4) and t = 10? Explain using mathematical evidence. (3 points)

Answer 1

Part A

`(x-r)^2+(y-s)^2=t^2`

where:

• (r, s) is the center of the circle.
• (x, y) is a point on the circle.
• t is the radius of the circle.

Part B

Domain = [-3, 17]

Range = [-14, 6]

Part C

Point (9, 1) is inside the border of the circle.

Explanation:

Part A

`\boxed{\begin{minipage}{9 cm}\underline{Pythagoras Theorem} \\\\$a^2+b^2=c^2$\\\\where:\\ \phantom{ww}$\bullet$ $a$ and $b$ are the legs of the right triangle. \\ \phantom{ww}$\bullet$ $c$ is the hypotenuse (longest side) of the right triangle.\\\end{minipage}}`

From inspection of the given triangle:

• a = x - r
• b = y - s
• c = t

Substitute these values into the formula to derive the standard equation of the circle:

`\boxed{ (x-r)^2+(y-s)^2=t^2}`

where:

• (r, s) is the center of the circle.
• (x, y) is a point on the circle.
• t is the radius of the circle.

Part B

Given the center of the circle (r, s) is (7, -4) and the radius (t) is 10.

The domain of the circle is the x-value of the center minus and plus the radius:

`\begin{aligned}\implies \textsf{Domain}&=[r-t,r+t]\\&= [7-10, 7+10] \\&= [-3, 17]\end{aligned}`

The range of the circle is the y-value of the center minus and plus the radius:

`\begin{aligned}\implies \textsf{Range}&=[s-t,s+t]\\&=[-4-10, -4+10]\\& = [-14, 6]\end{aligned}`

Part C

Substitute (r, s) = (7, –4) and t = 10 into the derived equation from part A:

`\implies (x-r)^2+(y-s)^2=t^2`

`\implies (x-7)^2+(y-(-4))^2=10^2`

`\implies (x-7)^2+(y+4)^2=100`

Substitute the given point (9, 1) into the equation:

`\begin{aligned}\implies (9-7)^2+(1+4)^2&=2^2+5^2\\&=4+25\\&=29\end{aligned}`

As 29 < 100, the point (9, 1) is inside the border of the circle.