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# Engineering

Finds the least distance of the point (3,16) from the circle ^{x^2{}}+ ^{y^2{}}+4x-8y-5=0

Answers (1)
S Sayak

The equation of the given circle is

\\x^{2}+y^{2}+4x-8y-5=0\\ x^{2}+4x+4+y^{2}-8y+16=4+16+5\\ (x+2)^{2}+(y-4)^{2}=5^{2}

The given circle has its centre at (-2,4) and has radius (r) of 5 units

Distance between the given point and centre of the circle is d

\\d=\sqrt{(3-(-2))^{2}+(16-4)^{2}}\\ d=13

The least distance of the point (3,16) from the circle ^{x^2{}}+ ^{y^2{}}+4x-8y-5=0 is d - r = 13 - 5 = 8 units

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