Question

in the xy - plane, which of the following points lies on a circle with the equation (x + k)^2+(y - j)^2 = 841?
a. (k + 20,j - 21)
b. (k - 20,j - 21)
c. (-k + 21,j + 20)
d. (-k - 21,-j - 20)

Explanation:

Step1: Recall circle - equation property

The equation of a circle is \((x - a)^2+(y - b)^2 = r^2\), where \((a,b)\) is the center of the circle and \(r\) is the radius. In the given equation \((x + k)^2+(y - j)^2=841\), the center of the circle is \((-k,j)\) and the radius \(r=\sqrt{841} = 29\).

Step2: Check each point

For a point \((x_0,y_0)\) to lie on the circle, it must satisfy the equation \((x_0 + k)^2+(y_0 - j)^2=841\).
Let's check option - a:
If \(x=k + 20\) and \(y=j+21\), then \((x + k)^2+(y - j)^2=((k + 20)+k)^2+((j + 21)-j)^2=(2k + 20)^2+21^2
eq841\) in general.
Let's check option - b:
If \(x=k - 20\) and \(y=j-21\), then \((x + k)^2+(y - j)^2=((k - 20)+k)^2+((j - 21)-j)^2=(2k - 20)^2+(-21)^2
eq841\) in general.
Let's check option - c:
If \(x=-k + 21\) and \(y=j + 20\), then \((x + k)^2+(y - j)^2=((-k + 21)+k)^2+((j + 20)-j)^2=21^2+20^2\).
We know that \(21^2=441\) and \(20^2 = 400\), so \(21^2+20^2=441 + 400=841\).
Let's check option - d:
If \(x=-k - 21\) and \(y=-j-20\), then \((x + k)^2+(y - j)^2=((-k - 21)+k)^2+((-j - 20)-j)^2=(-21)^2+(-2j - 20)^2
eq841\) in general.

Answer:

c. \((-k + 21,j + 20)\)