How do you find the equation that represents the image of circle (x- 5)^2 + (y + 12)^2 = 169(x−5)2+(y+12)2=169 after a translation 2 units right and 3 units down?

1 Answer
George C.
Jan 30, 2016

Adjust to standard form to see what the original centre is, translate the centre, then plug back into the equation to find:

(x-7)^2+(y-(-15))^2 = 13^2(x−7)2+(y−(−15))2=132

or:

(x-7)^2+(y+15)^2=169(x−7)2+(y+15)2=169

Explanation:

The standard equation of a circle with centre (h,k)(h,k) and radius rr is:

(x-h)^2+(y-k)^2 = r^2(x−h)2+(y−k)2=r2

With minor changes our starting equation is:

(x-5)^2+(y-(-12))^2=13^2(x−5)2+(y−(−12))2=132

That is: It is a circle with centre (5, -12)(5,−12) and radius 1313.

If translated 22 units right and 33 units down then the centre will be (5+2, -12-3) = (7, -15)(5+2,−12−3)=(7,−15) and the equation becomes:

(x-7)^2+(y-(-15))^2 = 13^2(x−7)2+(y−(−15))2=132

or putting it back in similar form to the starting equation:

(x-7)^2+(y+15)^2=169(x−7)2+(y+15)2=169

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