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Question

Find the equation to the tangent to the circle x2+y2=a2 which
(i) is parallel to the straight line y=mx+c,
(ii) is perpendicular to the straight line y=mx+c,
(iii) passes through the point (b,0),
and (iv) makes with the axes a triangle whose area is a^{2}$.

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Solution

(i) Let the straight line be y=mx+c1
Since it is tangent to the circle x2+y2=a2
So, a=|0+0c1|1+m2
c1=a(1+m2) or, c1=a(1+m2)
Hence, the tangents are
y=mx+a(1+m2) and y=mxa(1+m2)

(ii) Let the straight line be y=m1x+c2
m1=1m
Since it is tangent to the circle x2+y2=a2
So, a=|0+0c2|1+1m2
c2=a(1+m2)m or c2=a(1+m2)m
So, the equation of tangent are
y=1mx+a(1+m2)m and y=1mxa(1+m2)m

(iii) Let the straight line be y=mx+c
The point passes through (b,0)
So, 0=bm+c
c=bm
Since it is the tangent to the circle x2+y2=a2
So,a=|0+0+bm|1+m2
Solving we get the tangent as
y=a2b2a2xbm and y=a2b2a2xbm

(iv) Let the straight line be y=mx+c
It cuts x-axis at (cm,0) and y-axis at (0,c).
Area of triangle is a2 which is also equal to 12.cm.c.
c22m=a2
Since it is the tangent to the circle x2+y2=a2
So, a=|0+0c|1+m2
On solving we get,
m=1 and c=2a
So, the tangent is, y=x+2a.

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