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3 years ago

Write the equation of the circle centered at (0,1) with radius 15.

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

6 0

Answer:

$x^2+(y-1)^2=225$

Step-by-step explanation:

we know that

The equation of the circle in center radius form is equal to

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

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$(h,k)=(0,1)\\r=15\ units$

substitute

$(x-0)^2+(y-1)^2=15^2$

$x^2+(y-1)^2=225$

You might be interested in

Which strategie would eliminate a variable in the system of equations? −x+6y=8 7x−y=−2

NeX [460]

Answer:

substitution (or addition)

Step-by-step explanation:

A simple strategy for this system is to use substitution. The first equation is easily solved for x, so you could substitute that into the second equation:

x = 6y -8

7(6y -8) -y = -2 . . . . . x variable eliminated

The second equation is easily solved for y, so you could substitute that into the first equation.

y = 7x +2

-x +6(7x +2) = 8 . . . . . y-variable eliminated

The "addition" method is always a good way to eliminate a variable.

When the coefficient of a variable in one equation is a divisor of the coefficient of that variable in the other equation, a simple multiplication and addition will do.

To make the coefficient of x in the first equation the opposite of the coefficient of x in the second, multiply the first equation by 7. Adding that result to the second equation will eliminate x:

7(-x +6y) +(7x -y) = 7(8) +(-2)

42y -y = 56 -2 . . . . . . x-variable eliminated

Likewise, the second equation can be multiplied by 6 and added to the first to eliminate the y-variable:

(-x +6y) +6(7x -y) = (8) +6(-2)

-x +42x = -4 . . . . . . . . y-variable eliminated

It is often the case that using either substitution or "addition" requires about the same amount of work.

Here, the solutions are (x, y) = (-4/41, 54/41).

3 0

3 years ago

Which equation represents the circle described?

bazaltina [42]

We have the equation:

$x^2+y^2-8x-6y+24=0$

By arranging this equation in terms of x and y, we have:

$x^2-8x+y^2-6y=-24 \\ \\$

By using the method of completing the square, we have:

$x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}$

The center of this circle is:

$(h,k)=(4,3)$

So the equation that fulfills the statement is:

$(x-4)^2+(y-3)^2=2^2$

Finally, the right answer is c)

6 0

3 years ago

Read 2 more answers

Point b is the midpoint of ac-. if ab= 4x+2 and ac=10x-6, find the length of ac

Natasha2012 [34]

let's bear in mind that B is the midpoint and thus it cuts a segment into two equal halves.

$\bf \underset{\leftarrow \qquad \textit{\large 10x-6}\qquad \to }{\boxed{A}\stackrel{4x+2}{\rule[0.35em]{10em}{0.25pt}} B\stackrel{\underline{4x+2}}{\rule[0.35em]{10em}{0.25pt}\boxed{C}}} \\\\\\ AC=AB+BC\implies 10x-6=(4x+2)+(4x+2)\implies 10x-6=8x+4 \\\\\\ 2x-6=4\implies 2x=10\implies x=\cfrac{10}{2}\implies x= 5 \\\\[-0.35em] ~\dotfill\\\\ AC=(4x+2)+(4x+2)\implies AC=[4(5)+2]+[4(5)+2] \\\\\\ AC=22+22\implies AC=44$