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

BALAJI SANKARAN , 10 Years ago
Grade 12th pass
anser 2 Answers
Trina
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...
So 
r/2=|x-y-2|/√2 at x=0=y
r/2=2/√2
r=4/√2=2√2
 
Last Activity: 10 Years ago
BALAJI SANKARAN
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)
                               :(y-0)  = -1(x-0)
                               :     y  =  -x….…(ii)
from (i) & (ii)
 (x, y) = (1,-1) which is a point on the circle
 using distance formula on (0,0) & (1,-1)
we get the value of the radius =  \sqrt{}2
            
         
Last Activity: 10 Years ago
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