Answer:
Step-by-step explanation:
We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5).
First, recall that the equation of a circle is given by:
Where (<em>h, k</em>) is the center and <em>r</em> is the radius.
Since our center is at (7, 2), <em>h</em> = 7 and <em>k</em> = 2. Substitute:
Next, the since a point on the circle is (2, 5), <em>y</em> = 5 when <em>x</em> = 2. Substitute:
Solve for <em>r: </em>
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Simplify. Thus:
Finally, add:
We don't need to take the square root of both sides, as we will have the square it again anyways.
Therefore, our equation is:
Given that the triangle is a right angled triangle, to solve for x we shall use the Pythagorean theorem given by:
c^2=a^2+b^2
where:
c is the hypotenuse
a and b are the legs.
c=10
a=6
b=(x-6)
thus plugging the values in the equation and solving for x we get:
10^2=6^2+(x-6)^2
100=36+x^2-12x+36
100=x^2-12x+72
0=x^2-12x-28
factoring the above we get:
0=(x+2)(x-14)
hence
x=-2 or x=14
given that there is no negative distance then
x=14
thus the value of x is 14
There could be five in each class.
I don't know what you are talking about please re word the question its really hard to understand what your asking.