Circle Ques 26
- The lines $2 x-3 y=5$ and $3 x-4 y=7$ are diameters of a circle of area $154 sq$ units. Then, the equation of this circle is
$(1989,2 M)$
(a) $x^{2}+y^{2}+2 x-2 y=62$
(b) $x^{2}+y^{2}+2 x-2 y=47$
(c) $x^{2}+y^{2}-2 x+2 y=47$
(d) $x^{2}+y^{2}-2 x+2 y=62$
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Answer:
Correct Answer: 26.(c)
Solution:
Formula:
- Since, $2 x-3 y=5$ and $3 x-4 y=7$ are diameters of a circle.
Their point of intersection is centre $(1,-1)$.
Also given,
$ \pi r^{2}=154 $
$ \Rightarrow \quad r^{2}=154 \times \frac{7}{22} \Rightarrow r=7 $
$\therefore$ Required equation of circle is
$ \begin{array}{ll} & (x-1)^{2}+(y+1)^{2}=7^{2} \\ \Rightarrow \quad & x^{2}+y^{2}-2 x+2 y=47 \end{array} $
BETA
