SajjadAhmad wrote:

In the figure to the right, the circle is centered at the origin and passes through point P. Which of the following points does it also pass through?

(A) (3, 3)

(B) (-2 \(\sqrt{2}\), -1)

(C) (2, 6)

(D) (-3 \(\sqrt{3}\), \(\sqrt{3}\))

(E) (–3, 4)

Nova GMAT

I used the equation of the circle. For a circle to pass through a point, that point's coordinates must satisfy the circle's equation.

General formula: \((x - h)^2 + (y - k)^2 = r^2\)

With center at origin, substitute (0,0) for x and y; and from point (0, -3), substitute those numbers respectively for h and k to find r.

(0 - 0)\(^2\) + (0 - -3)\(^2\) = r\(^2\)

= \(0 + 9 = r^2\)

\(r = 3\)

So the circle's equation is \(x^2 + y^2 = 3^2\), or

\(x^2 + y^2 = 9\)

Answers - A, C, D, and E can all be ruled out quickly. If, as is the case with A and E, a point has one coordinate whose absolute value is 3, to satisfy this equation, the other coordinate would have to be 0, because both \((-3)^2\) and \(3^2\) \(= 9\):

\((-3)^2 + y^2 = 9\)

\(9 + y^2 = 9\)

\(y^2 = 0\)

y must equal = 0. Reject A and E.

And if, as is the case with C and D, one coordinate's absolute value is greater than 3 (6, and -3 \(\sqrt{3}\), respectively), its squared value will certainly be greater than 9, let alone when added to the square of the other coordinate's absolute value.

So (B), with coordinates \((-2\sqrt{2}, -1)\):

\((-2\sqrt{2})^2\) + \((-1)^2\) =

\(8 + 1 = 9.\) Correct.

_________________

At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123