First slide

Definition of a circle

Question

If the lines $3 x - 4 y - 7 = 0$ and $2 x - 3 y - 5 = 0$ are diameters of a circle of area $49 π$ square units, the equation of the circle, is

Moderate

A

$x^{2} + y^{2} + 2 x - 2 y - 62 = 0$
B

$x^{2} + y^{2} - 2 x + 2 y - 62 = 0$
C

$x^{2} + y^{2} - 2 x + 2 y - 47 = 0$
D

$x^{2} + y^{2} + 2 x - 2 y - 47 = 0$

Solution

The centre of the circle is the point of intersection of its
diameters $3 x - 4 y - 7 = 0$ and $2 x - 3 y - 5 = 0$ So, its coordinates are $(1 - 1)$ .
Let $r$ be the radius of the circle. Then
$π r^{2} = 49 π \Rightarrow r = 7$
Hence, the equation of the circle is
$(x - 1)^{2} + (y + 1)^{2} = 7^{2}$ or , $x^{2} + y^{2} - 2 x - 2 y - 47 = 0$

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