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A straight line x=y+2 touches the circle (x^2 + y^2) = r^2. The value of r is

A straight line x=y+2 touches the circle (x^2 + y^2) = r^2. The value of r is

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Grade:12th pass

2 Answers

Trina
34 Points
8 years ago
We have the equation of the circle as 4( x^2+y^2)=r^2
X^2+y^2=(r/2)^2
Now x=y+2 is a tangent to the circle....
Centre of the circle is (0,0)
Tangent is always perpendicular to the radius...
Soย 
r/2=|x-y-2|/โˆš2 at x=0=y
r/2=2/โˆš2
r=4/โˆš2=2โˆš2
ย 
BALAJI SANKARAN
10 Points
8 years ago
Equation of circle :4(x2 +y2)=r2
centre of the circle = (0, 0)
equation of tangent :x=y+2 [y=x-2].....(i)
slope of the tangent = 1
slope of radius = -1
using one-point form :(y-y1) = m(x-x1)
ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  :(y-0)ย  = -1(x-0)
ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  :ย ย ย ย  yย  =ย  -xโ€ฆ.โ€ฆ(ii)
from (i) & (ii)
ย (x, y) = (1,-1) which is a point on the circle
ย using distance formula on (0,0) & (1,-1)
we get the value of the radius =ย  \sqrt{}2
ย ย ย ย ย ย ย ย ย ย ย ย 
ย ย ย ย ย ย ย ย ย 

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