What does (x + 2) & sup2; + Y & sup2; = 5 say about the equation of a circle symmetrical at the origin

(x-2)²+y²=5

Given the equation (x-3) & sup2; + (y-4) & sup2; = 4 of a circle, find the slope of a line passing through the origin of the coordinate and tangent to the circle?

Center (3,4), radius 2
Over origin
Let the line be kx-y = 0
The distance from the center of the circle to the tangent is equal to the radius
So | 3k-4 | / √ (K & sup2; + 1) = 2
square
9k²-24k+16=4k²+4
5k²-24k+12=0
So k = (12 ± √ 69) / 5

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