y=3x+c is a tangent to the circle x^2+y^2-2x-4y -5=0 , then c is =?
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given equation of circle is x^2+y^2-2x-4y+5 = 0
equation of tangent is 3x-y+c = 0
now the distance of the tangent from the centre of the circle= radius of circle
for a circle x^2 +y^2 +2gx +2fy +c=0 , the centre is (-g,-f) and radius is (f^2 +g^2-c)^1/2
distance of a line(ax+by+c=0) from a point (x1,y1) is | ax1+by1+c | / (a^2+b^2)^1/2
therefore just substitute the value of the centre (1,2) of the circle in the above equation and equate it to the radius (10)^1/2
we get |1+c|=10
implies c=+9 or c=-11
equation of tangent is 3x-y+c = 0
now the distance of the tangent from the centre of the circle= radius of circle
for a circle x^2 +y^2 +2gx +2fy +c=0 , the centre is (-g,-f) and radius is (f^2 +g^2-c)^1/2
distance of a line(ax+by+c=0) from a point (x1,y1) is | ax1+by1+c | / (a^2+b^2)^1/2
therefore just substitute the value of the centre (1,2) of the circle in the above equation and equate it to the radius (10)^1/2
we get |1+c|=10
implies c=+9 or c=-11
Nirmal11:
C=9
WHY IS -11 NOT THE SOLUTION?
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