Question

(x-6)^2+(y+5)^2=16

In the xy-plane, the graph of the equation above is a

circle. Point P is on the circle and has coordinates

10, −5 . If PQ is a diameter of the circle, what are

the coordinates of point Q ?

A) 2, −5

B) 6, −1

C) 6, −5

D) 6,-9​

Answer 1

Answer:

(2,-5)

Step-by-step explanation:

$(x - 6) {}^{2} + {(y + 5)}^{2} = 16$

This given equation is an equation of a circle.

Now, On comparing it with :

${(x - h)}^{2} + {(y - k)}^{2} = r {}^{2}$

Where (h,k) are the coordinates of the center O of the circle which are : (6,-5)

Given : Coordinates of point P (10,-5)

Let the point Q be (x,y)

So, Point O is the mid point of PQ. Now, Applying the mid point Formula:

$( \frac{x1 + x2}{2}) ( \frac{y1 + y2}{2} )$

$for \: \: x \: \: coordinate \: \: : \ \\ ( \frac{10 + x}{2} ) = 6 \\ 10 + x = 2 \times 6 \\ 10 + x = 12 \\ x = 2$

$for \: \: y \: \: coordinate \: : \\ ( \frac{ - 5 + y}{2} ) = - 5 \\ - 5 + y = - 5 \times 2 \\ - 5 + y = - 10 \\ y = - 5$

So, the Coordinates of Point Q are

(2,-5)