88.2k views
2 votes
How do you solve this?

How do you solve this?-example-1

2 Answers

3 votes

The equation of a circle is:

(x - h)² + (y - k)² = radius²

Note:

h = x coordinate for the centre of the circle

k = y coordinate for the centre of the circle

The equation for the circle in the question is:

(x - 6)² + (y - 5)² = 16

So the coordinates for the centre of the circle is:

(6, 5)

------------------------

Since PQ is the diameter of the circle, that means the coordinates for the midpoint of PQ would also be the coordinates for the centre of the circle

That means:

( (x coords of P and Q) / 2 , (y coords of P and Q) / 2 )= (6 , 5)

So x-coords of Q:


(10 + x)/(2) = 6

10 + x = 12

x = 2

y-coords of Q


(-5+ x)/(2) = -5

-5 + x = - 10

x = 5

So the coordinates for Q are:

(2, -5)

---------------------------------------------

Answer:

Option A) (2, - 5)

User Potasmic
by
8.1k points
4 votes


\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] ~\dotfill\\\\ (x-6)^2+(y+5)^2=16\implies [x-\stackrel{h}{6}]^2+[y-(\stackrel{k}{-5})]^2=\stackrel{r}{4^2}~\hfill \begin{cases} \stackrel{center}{(6,-5)}\\ \stackrel{radius}{4} \end{cases}

Check the picture below.

How do you solve this?-example-1
User Bhansa
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories