Answers:
Question 51. [tex]\boldsymbol{(x-3)^2 + (y-1)^2 = 4}[/tex]
Question 52. [tex]\boldsymbol{(x+3)^2 + (y-2)^2 = 4}[/tex]
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Explanation:
The template of any circle is
[tex](x-h)^2 + (y-k)^2 = r^2[/tex]
with (h,k) being the center and r as the radius
For question 51, the center is (3, 1) and the radius is 2. So,
[tex](x-h)^2 + (y-k)^2 = r^2\\\\(x-3)^2 + (y-1)^2 = 2^2\\\\(x-3)^2 + (y-1)^2 = 4[/tex]
Question 52 will have similar steps. This time the center is (-3, 2) and the radius is 2.
