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the circle x^2+y^2-8x+4y+4=0 touches a)x-axis b)y-axis c)both axes d)neither ?? i evaluated by putting values of x and y as zero and got real values which states it touches both axes??dosent it? but solution says b)y-axis

the circle x^2+y^2-8x+4y+4=0 touches
a)x-axis
b)y-axis
c)both axes
d)neither
??
i evaluated by putting values of x and y as zero and got real values which states it touches both axes??dosent it?
but solution says b)y-axis

Grade:11

2 Answers

vikas askiitian expert
509 Points
11 years ago

dear anmol

on putting x=0 , we have

y2+4y+4=0

(y+2)2 = 0

y=-2                only 1 value of y that means it touches y axis ...

now at y=0 , we have

x2-8x+4 = 0

here we have 2 reaL values of x , that means it cuts the x axis at two points ..

so option b is correct

Durgesh Prasher
21 Points
11 years ago

DO YOU KNOW THAT COORDS OF CENTRE OF CIRCLE ARE {-1/2 COEFF. OF X ,-1/2  COEFF. OF Y}

RADIUS OF CIRCLE [X1,Y1] IS UNDER ROOT OF X1^2 +Y1^2 -C

FOR THIS EQUATION COORDS ARE [4,-2]

N RADIUS IS 4

SO IT WILL TOUCH Y AXIS AS ABSICCA =RADIUS

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