Question

completing the square & distance
practice questions
r on the following questions.
$(x - 2)^2+(y + 5)^2 = 25$

1. in the $xy$-plane, the graph of the equation

above is a circle. point $g$ is on the circle and
has coordinates $(7, - 5)$. if $overline{fg}$ is a diameter
of the circle, what are the coordinates of point
$f$?
(a) $(-3, - 5)$
(b) $(2,5)$
(c) $(5,0)$
(d) $(5, - 10)$

1. a circle in the $xy$-plane has equation

$(x - 4)^2+(y + 3)^2 = 36$. which of the followin
points does not lie in the interior of the

Explanation:

Step1: Find the center of the circle

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center. For the equation \((x - 2)^2+(y + 5)^2=25\), the center \(C\) has coordinates \((2,-5)\).

Step2: Use the mid - point formula

The mid - point formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). Let \(F=(x,y)\) and \(G=(7,-5)\), and the mid - point (center of the circle) \((2,-5)\).
For the \(x\) - coordinate: \(\frac{x + 7}{2}=2\), which gives \(x+7 = 4\), so \(x=-3\).
For the \(y\) - coordinate: \(\frac{y+( - 5)}{2}=-5\), which gives \(y-5=-10\), so \(y=-5\).

Answer:

A. \((-3,-5)\)