Question:
The lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of a circle of area 154 sq unitis. Then, the equation of this circle is
The lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of a circle of area 154 sq unitis. Then, the equation of this circle is
Updated On: Jun 14, 2022
- $x^2 + y^2 + 2x - 2y = 62$
- $x^2 + y^2 + 2x - 2y = 47$
- $x^2 + y^2 - 2x + 2y =47$
- $x^2 + y^2 - 2x + 2y = 62$
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The Correct Option is C
Solution and Explanation
Since, 2x - 3y = 5 and 3x - 4y = 7 are diam eters of a
circle.
Their point of intersection is centre (1, - 1 ) .
Also given, $\pi r^2$ =154
$\Rightarrow$ $r^2=154 \times \frac {7}{22}$
$\Rightarrow r=7$
$\therefore$ Required equation of circle is
$(x - 1)^2 + (y + 1)^2 = 7^2$
$\Rightarrow$ $x^2 + y^2 - 2x + 2y = 47$
circle.
Their point of intersection is centre (1, - 1 ) .
Also given, $\pi r^2$ =154
$\Rightarrow$ $r^2=154 \times \frac {7}{22}$
$\Rightarrow r=7$
$\therefore$ Required equation of circle is
$(x - 1)^2 + (y + 1)^2 = 7^2$
$\Rightarrow$ $x^2 + y^2 - 2x + 2y = 47$
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