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10-1.Circle and System of Circles
hard
The equation of three circles are ${x^2} + {y^2} - 12x - 16y + 64 = 0,$ $3{x^2} + 3{y^2} - 36x + 81 = 0$ and ${x^2} + {y^2} - 16x + 81 = 0.$ The co-ordinates of the point from which the length of tangent drawn to each of the three circle is equal is
A
$(33/4, 2)$
B
$(2, 2)$
C
$(2, 33/4)$
D
None of these
Solution
(d) The required point is the radical centre of the three given circles.
Now, ${S_1} – {S_2} = 0 \Rightarrow – 16y + 37 = 0$
${S_2} – {S_3} = 0\, \Rightarrow \,4x – 54 = 0$
${S_3} – {S_1} = 0\, \Rightarrow \, – 4x + 16y + 17 = 0$
Solving these equations, we get
$x = \frac{{54}}{4},\,\,y = \frac{{37}}{{16}}$$ \Rightarrow $$x = \frac{{27}}{2},\,y = \frac{{37}}{{16}}$
Hence the required point is $\left( {\frac{{27}}{2},\,\frac{{37}}{{16}}} \right)$.
Standard 11
Mathematics
