Circle Ques 63
- Two circles, each of radius 5 units, touch each other at $(1,2)$. If the equation of their common tangent is $4 x+3 y=10$, find the equations of the circles. $(1991,4 M)$
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Answer:
63.($x-5)^2$ + ($y-5)^2$ = $5^2$ and ($x+3)^2$ + ($y+1)^2$ = $5^2$
Solution:
Formula:
- We have,
Slope of the common tangent $=-\frac{4}{3}$
$ \therefore \quad \text { Slope of } C _1 C _2=\frac{3}{4} $
If $C_1 C_2$ makes an angle $\theta$ with $X$-axis, then $\cos \theta=\frac{4}{5}$ and $\sin \theta=\frac{3}{5}$.
So, the equation of $C_1 C_2$ in parametric form is
$ \frac{x-1}{4 / 5}=\frac{y-2}{3 / 5} ……(i) $
Since $C_1$ and $C_2$ are points on Eq. (i) at a distance of 5 units from $P$.
So, the coordinates of $C_1$ and $C_2$ are given by
$ \frac{x-1}{4 / 5}=\frac{y-2}{3 / 5}= \pm 5 \Rightarrow x=1 \pm 2 $
$ \text { and } \quad y=2 \pm 3 \text {. } $
Thus, the coordinates of $C _1$ and $C _2$ are $(5,5)$ and $(-3,-1)$, respectively.
Hence, the equations of the two circles are
and
$ \begin{alignedat} & (x-5)^{2}+(y-5)^{2}=5^{2} \\ & (x+3)^{2}+(y+1)^{2}=5^{2} \end{aligned} $
BETA
