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3 years ago

The points J (-13,10) and K(-1,-6) are endpoints of a diameter of circle G. What is the equation of circle G?

2 answers:

7 0

The correct answer is D

The equation of a circle is (x-h)^2 + (y-k)^2 = r^2

to find (h,k), you find the middle of the circle, in this scenario you do so by finding the middle of the diameter, a line that goes through the center of the circle.

To find the X value of the midpoint, add both x values together and divide by 2 and repeat for y

-13 + -1 = -14
-14/2 =-7

10+ -6 = 4
4/2 = 2

therefore (h,k) = ( -7, 2 )

Next plug these values in the equation of a circle

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

becomes

(x- (-7)) ^2 + (y-(2)) ^2 = r^2

to find r, use the distance formula to find the length of the diameter, 20, and divide by 2

plug 10 in for r and you get 100

(x+7)^2 + (y-2)^2 = 100

sorry for the late response

Alina [70]3 years ago

5 0

Answer:

(x + 7)2 + (y − 2)2 = 100

Step-by-step explanation:

I just solved this

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Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

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Answer:

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Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

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Here $-2$ that is $x$ satisfied. Then

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That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

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This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

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