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  • find the equation of diameter of the circle  which is perpendicular to&

find the equation of diameter of the circle x^2+y^2-4x-4y+4=0 which is perpendicular to y=-x+2 ?

Option: 1

y+x=4


Option: 2

y=x


Option: 3

y=x+2


Option: 4

None of these


Answers (1)

best_answer

 

 

DIAMETER OF A CIRCLE -

 DIAMETER OF A CIRCLE

The locus of the mid-points of a system of parallel chords of a circle is known as the diameter of the circle. 

The diameter of a circle always passes through the centre of a circle and perpendicular to the parallel chords

\\\mathrm{Let\;the\;equation\;of\;the\;circle\;be\;\;x^2+y^2=a^2\;\;and\;equation\;of}\\\mathrm{parallel\;chord\;AB\;is,\;\;y=mx+c.}

Equation of any diameter to the given circle is perpendicular to the given parallel chord is my + x + λ = 0 which passes through the centre (0, 0) of a circle.

m?0 + 0 +  λ = 0

 λ = 0

Hence, the required equation of the diameter is x + my = 0

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x^2+y^2-4x-4y+4=0\\ \text{centre of circle at (2,2,) and radius r=2}\\ \text{diameter is perpandicular to y=-x+2}\\ \text{so slope of the diamter =1 and passes through (2,2)}\\ \text{Equation of the diamter }\\ y-2=1(x-2)\\ y=x

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