DSQ: Calculating the equation of a circle - GMAT Quantitative

You cannot determine the equation of a circle knowing only the center, as in Statement 1.

But given the coordinates of any diameter of a circle, as in Statement 2, you can use the midpoint formula to find the center, and the distance formula to find the distance from the center to either endpoint - this is the radius. Once you know the center and the radius, just apply them to the standard form of the equation of a circle.

The center of the circle of the equation is the point .

If Statement 2 is assumed, then ; since is known to be positive, is negative, which is the same as Statement 1.

From either statement, we know to be negative, and, subsequently, that both coordinates of the center are negative, putting it in Quadrant III.

The equation of a circle is given by:

Where r is the radius and (h,k) are the center of the circle

I) gives us (h,k)

By using II) and distance formula, we can find r

Thus, both statements are needed!

To write the equation of a circle we need the coordinates of its center and the length of its radius.

I) Gives us the center.

II) Tells us that the circle just touches the x-axis at the point (7,0), which is exactly 7 units below our center. This makes our radius 7.

Thus, we need both statements.

The equation of a circle is as follows:

Where is the center of the circle and is the radius.

Statement I tells us how to relate and , but doesn't tell us how to find either one.

Statement II gives us .

We almost have enough information here, but we can't tell for certain where the circle's center is, so more information is needed.

The equation of a circle is as follows:

Where is the center of the circle and is the radius.

Statement I gives us the radius.

Statement II gives us clues to find .

So, by using both, we can find the equation. Both are needed.

Recap:

Consider circle L.

I) The radius of L is twice the value of the third prime number.

II) The x-coordinate of L is four times the y-coordinate of L. The y-coordinate is equal to the radius.

Find the equation of circle L.

Using Statement I, the third prime number is 5, so our radius is 10.

Using Statement II, , so must be 40.

Putting it all into the equation of a circle:

The equation of a circle is as follows:

Where is the center of the circle and is the radius.

Statement I gives us a clue to find the diameter, which can be used to find the radius.

Statement II gives us the center of the circle.

Use Statement II and Statement I to find the radius, then plug it all into the equation to find the equation of Circle T.

Recap:

Circle T is plotted in Euclidean space. Find its equation.

I) The diameter of circle T is twice the square of the y-coordinate of its center.

II) Circle T is centered at the point .

Statement II gives us .

Use Statement I and Statement II to write the following equation:

So, if the diameter is 18, the radius must be 9. This makes our equation of the circle:

Circle equations are written in the form where are the coordinates for the center of the circle. When dealing with the area of a circle, the formula is Since, the circle equation already gives us the squared radius, we only need to multiply that number by Therefore, the area is

You cannot determine the equation of a circle knowing only the center, as in Statement 1.

But given the coordinates of any diameter of a circle, as in Statement 2, you can use the midpoint formula to find the center, and the distance formula to find the distance from the center to either endpoint - this is the radius. Once you know the center and the radius, just apply them to the standard form of the equation of a circle.

The center of the circle of the equation is the point .

If Statement 2 is assumed, then ; since is known to be positive, is negative, which is the same as Statement 1.

From either statement, we know to be negative, and, subsequently, that both coordinates of the center are negative, putting it in Quadrant III.

The equation of a circle is given by:

Where r is the radius and (h,k) are the center of the circle

I) gives us (h,k)

By using II) and distance formula, we can find r

Thus, both statements are needed!

To write the equation of a circle we need the coordinates of its center and the length of its radius.

I) Gives us the center.

II) Tells us that the circle just touches the x-axis at the point (7,0), which is exactly 7 units below our center. This makes our radius 7.

Thus, we need both statements.

The equation of a circle is as follows:

Where is the center of the circle and is the radius.

Statement I tells us how to relate and , but doesn't tell us how to find either one.

Statement II gives us .

We almost have enough information here, but we can't tell for certain where the circle's center is, so more information is needed.

The equation of a circle is as follows:

Where is the center of the circle and is the radius.

Statement I gives us the radius.

Statement II gives us clues to find .

So, by using both, we can find the equation. Both are needed.

Recap:

Consider circle L.

I) The radius of L is twice the value of the third prime number.

II) The x-coordinate of L is four times the y-coordinate of L. The y-coordinate is equal to the radius.

Find the equation of circle L.

Using Statement I, the third prime number is 5, so our radius is 10.

Using Statement II, , so must be 40.

Putting it all into the equation of a circle:

The equation of a circle is as follows:

Where is the center of the circle and is the radius.

Statement I gives us a clue to find the diameter, which can be used to find the radius.

Statement II gives us the center of the circle.

Use Statement II and Statement I to find the radius, then plug it all into the equation to find the equation of Circle T.

Recap:

Circle T is plotted in Euclidean space. Find its equation.

I) The diameter of circle T is twice the square of the y-coordinate of its center.

II) Circle T is centered at the point .

Statement II gives us .

Use Statement I and Statement II to write the following equation:

So, if the diameter is 18, the radius must be 9. This makes our equation of the circle:

Circle equations are written in the form where are the coordinates for the center of the circle. When dealing with the area of a circle, the formula is Since, the circle equation already gives us the squared radius, we only need to multiply that number by Therefore, the area is

You cannot determine the equation of a circle knowing only the center, as in Statement 1.

But given the coordinates of any diameter of a circle, as in Statement 2, you can use the midpoint formula to find the center, and the distance formula to find the distance from the center to either endpoint - this is the radius. Once you know the center and the radius, just apply them to the standard form of the equation of a circle.

The center of the circle of the equation is the point .

If Statement 2 is assumed, then ; since is known to be positive, is negative, which is the same as Statement 1.

From either statement, we know to be negative, and, subsequently, that both coordinates of the center are negative, putting it in Quadrant III.

The equation of a circle is given by:

Where r is the radius and (h,k) are the center of the circle

I) gives us (h,k)

By using II) and distance formula, we can find r

Thus, both statements are needed!

To write the equation of a circle we need the coordinates of its center and the length of its radius.

I) Gives us the center.

II) Tells us that the circle just touches the x-axis at the point (7,0), which is exactly 7 units below our center. This makes our radius 7.

Thus, we need both statements.