Problem from practice SAT II Math II test

[lyzhou1990]

I was taking a practice test when I got up to the following question and was completely stumped. Could someone tell me the answer (and how to get it?) Thanks.



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[stapel]

To tutors: The image (too small to be viewed here) contains the following:

Which of the following can be the equation of the circle given in Figure 3?

. . . . .A. x² + y² - 14x - 6y = 0
. . . . .B. x² + y² - 14x - 6y + 58 = 0
. . . . .C. x² + y² - 6x - 14y + 58 = 0
. . . . .D. x² + y² + 6x - 14y + 42 = 0
. . . . .E. x² + y² - 6x - 14y + 42 = 0

Figure 3 shows the x- and y-axes, with a circle drawn in the first and second quadrants. The circle is almost entirely within the first quadrant. The circle intersects the y-axis at y = 4 and at y = 10. The circle does not touch or cross the x-axis.

To lyzhou1990: One method for solving this would be to plug the given points into the equations, and find which one works. Another method would be to use whatever you have learned in algebra and/or geometry to determine the circle equation, and then compare with the listed answers to select the correct one.

How far have you gotten? Where are you stuck?

Please reply showing all of your steps and reasoning. Thank you.

Eliz.

 

[soroban]

Hello, lyzhou1990!

I'm getting a "None of the above" answer . . .
. . Could the "42" be a typo?

            |
            |   * * *
     B(0,10)*           *
          * |             *
         *  |              *
            |
        *   |     C         *
        *   |     o         *
        *   |   (h,7)       *
            |
         *  |              *
          * |             *
      A(0,4)*           *
            |   * * *
            |
            |
      - - - + - - - - - - - - - -
            |

We have a circle with y-intecepts \(\displaystyle A(0,\,4)\) and \(\displaystyle B(0,\,10)\)

Its center is \(\displaystyle C(h,\,7)\)

Its radius is: \(\displaystyle \,r\:=\:\overline{CA} \:=\:\sqrt{h^2\,+\,9}\)

Hence, its equation is: \(\displaystyle \,(x\,-\,h)^2\,+\,(y\,-\,7)^2\:=\:h^2\,+\,9\)

Expand: \(\displaystyle \,x^2\,-\,2hx \,+\,h^2\,+\,y^2\,-\,14y\,+\,49\;=\;h^2\,+\,9\)

Simplify: \(\displaystyle \,x^2\,+\,y^2\,-\,2hx\,-\,14y\,+\,\)40\(\displaystyle \;=\;0\;\;\) . . . see?

 

[lyzhou1990]

Hi,
Thanks a lot for the help.

To stapel:
I got nowhere at all with this problem...I actually just skipped it, and then tried to figure out how to arrive at the answer after looking at the correct answer presented to the book. Sorry I can't be of more help.

To soroban:
I don't think it's a typo...or at least the book doesn't seem to imply that it is.

The book's brief and inadequate explanation simply stated that the formula for the circle was

(x-3)^2 + (y-7)^2 = 16

which fits the standard form of

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

I'm guessing that somehow it can be determined that h = 3 and that the radius = 4, since expanding the formula that the book gives as an answer puts answer choice E as the correct answer.

However, my major problem is that I have no idea how it can be determined that h = 3. Any help in this regard would be greatly appreciated!