Question

21 mark for review
$(x - 6)^2+(y + 3)^2 = 25$
the given equation represents circle ( p ) in the ( x y ) - plane. circle ( q ) has a center that is 3 units to the right of and 2 units below the center of circle ( p ). circle ( q ) has a diameter that is double the diameter of circle ( p ). which equation represents circle ( q )?
a ( (x - 9)^2+(y + 5)^2 = 100 )
b ( (x - 3)^2+(y + 1)^2 = 100 )
c ( (x - 9)^2+(y + 5)^2 = 50 )
d ( (x - 3)^2+(y + 1)^2 = 50 )

Explanation:

Step1: Find the center and radius of circle \(P\)

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.
For the equation \((x - 6)^2+(y + 3)^2 = 25\), we have \(h = 6\), \(k=-3\), and \(r=\sqrt{25}=5\) (since \(r^2 = 25\)).

Step2: Find the center of circle \(Q\)

The center of circle \(Q\) is \(3\) units to the right and \(2\) units below the center of circle \(P\).
Using the transformation \((h_Q,k_Q)=(h + 3,k-2)\).
Substitute \(h = 6\) and \(k=-3\): \(h_Q=6 + 3=9\), \(k_Q=-3-2=-5\).

Step3: Find the radius of circle \(Q\)

The diameter of circle \(Q\) is double the diameter of circle \(P\).
The diameter of circle \(P\) is \(d_P = 2r_P=10\), so the diameter of circle \(Q\) is \(d_Q=2\times10 = 20\), and the radius \(r_Q=\frac{d_Q}{2}=10\).

Step4: Write the equation of circle \(Q\)

Using the standard form \((x - h_Q)^2+(y - k_Q)^2=r_Q^2\).
Substitute \(h_Q = 9\), \(k_Q=-5\), and \(r_Q = 10\): \((x - 9)^2+(y+5)^2=10^2=100\).

Answer:

A. \((x - 9)^2+(y + 5)^2 = 100\)