Slope of tangent to x2+y2=1 at (21,21)
x2+y2=1
2x+2yy′=0
y′=−yx=−1
Given that, y=mx+c, is perpendicular to L1,
So, m=1⇒y=x+c
Now distance of (3,0) from y=x+c is
∣2c+3∣=1
c2+6c+9=2
c2+6c+7=0.
If a line y=mx+c, is a tangent to the circle (x−3)2+y2=1, and it is perpendicular to a line L1, where L1 is the tangent to the circle x2+y2=1, at the point (21,21), then
Held on 8 Jan 2020 · Verified 6 Jul 2026.
c2−7c+6=0
c2+7c+6=0
c2+6c+7=0
c2−6c+7=0
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