The distance of the point $(1, 2)$ from the centre of circle $x^{2} + y^{2} - 8 x + 4 y + 16 = 0$ is units

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Solution

Given equation of circle
$x^{2} + y^{2} - 8 x + 4 y + 16 = 0$
Centre $= (- g, - f) = (4, - 2)$
Distance of point $(1, 2)$ from the centre of circle
$= \sqrt{(4 - 1)^{2} + (- 2 - 2)^{2}}$
$= \sqrt{9 + 16} = 5 units$

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The distance of the point (1,2) from the centre of circle x2+y2−8x+4y+16=0 is units

Given equation of circle x2+y2−8x+4y+16=0 Centre =(−g,−f)=(4,−2) Distance of point (1,2) from the centre of circle =√(4−1)2+(−2−2)2 =√9+16=5 units

The distance of the point $(1, 2)$ from the centre of circle $x^{2} + y^{2} - 8 x + 4 y + 16 = 0$ is units

Given equation of circle
$x^{2} + y^{2} - 8 x + 4 y + 16 = 0$
Centre $= (- g, - f) = (4, - 2)$
Distance of point $(1, 2)$ from the centre of circle
$= \sqrt{(4 - 1)^{2} + (- 2 - 2)^{2}}$
$= \sqrt{9 + 16} = 5 units$