Question

The lines $2 \mathrm{x}-3 \mathrm{y}=5$ and $3 \mathrm{x}-4 \mathrm{y}=7$ are diameters of a circle having area as 154 sq. units. Then the equation of the circle is
A) ${{x}^{2}}+{{y}^{2}}+2x-2y=47$
B) $x^{2}+y^{2}-2 x+2 y=47$
C) ${{x}^{2}}+{{y}^{2}}+2x+2y=62$
D) $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-2 \mathrm{y}=62$

Answer 1

Hint: A circle's diameter is determined by multiplying the radius by two. The diameter is measured from one end of the circle to a point on the other end, passing through the centre, whereas the radius is measured from the centre of a circle to one endpoint on the circle's perimeter. The letter D is used to identify it. A circle has an infinite number of points on its circumference, which translates to an endless number of diameters with equal lengths for each diameter.

Complete Step by step solution:
The intersection of diagonals is at the centre.
$\Rightarrow 2 \mathrm{x}-3 \mathrm{y}=5$
$\Rightarrow 3 \mathrm{x}-4 \mathrm{y}=7$
Solving simultaneous equation, $x=1, y=-1$
Center $=(1,-1)$
The radius of a circle is the length of the straight line that connects the centre to any point on its circumference. Because a circle's circumference can contain an endless number of points, a circle can have more than one radius. This indicates that a circle has an endless number of radii and that each radius is equally spaced from the circle's centre. When the radius's length varies, the circle's size also changes.
Taking radius as $r$
$\Rightarrow \text{r}=\sqrt{\dfrac{154}{\pi }}$
$\Rightarrow \mathrm{r}=\sqrt{\dfrac{154}{\pi}} \Rightarrow \sqrt{\dfrac{154}{22} \times 7}=7$
Equation –
$\Rightarrow(x-1)^{2}+(y+1)^{2}=7^{2}$
$\Rightarrow x^{2}+1-2 x+y^{2}+1+2 y=49$
$\Rightarrow x^{2}+y^{2}-2 x+2 y=49-2$
$\Rightarrow x^{2}+y^{2}-2 x+2 y=47$
Hence, the correct option is (B).

Note:The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. The coordinates of the circle's centre and radius are found using this general form, where g, f, and c are constants. The general form of the equation of a circle makes it difficult to identify any significant properties about any specific circle, in contrast to the standard form, which is simpler to comprehend. So, to quickly change from the generic form to the standard form, we will use the completing the square formula.