the circle x^2+y^2-8x+4y+4=0 touchesa)x-axisb)y-axisc)both axesd)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

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

Approved

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