What is the point of symmetry for the circle (x – 5)2 + (y + 4)2 = 25?

In a circle, the point of symmetry is the centerpoint; any straight line that passes through this point is a line of symmetry. The standard form of equation of a circle is given as:

(x – h)^2 + (y – k )^2 = r^2

Where,

h = x coordinate of the center

k = y coordinate of the center

In the problem statement, we are given that the equation of the circle is:

(x – 5)^2 + (y + 4)^2 = 25

So identifying the variables from the standard form of equation of a circle:

h = 5

k = - 4

 

Therefore the point of symmetry is (5, -4).

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

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