NIMCET 2016 — Mathematics PYQ

The standard form of the equation of a circle is:

$$(\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{R}}^2$$

where(h,k) are the coordinates and the R is the radius of center of the circle

Area of the circle$=\pi {\mathrm{R}}^2$

Note: The intersection of the Equation of diameters is center of the circle

Calculation:

Given area of circle=154 sq.units

$\Rightarrow \pi$ ${\mathrm{R}}^2=154$

$\Rightarrow {\mathbb{R}}^2=154\ast \frac{7}{22}$

$\Rightarrow \mathbf{R}=7$

Equation of the diameters

$$2\mathrm{x}-3\mathrm{y}+12=0$$...(i)

$$\mathbf{x}+4\mathbf{y}-5=0$$...(ii)

Intersection of the diameters (i) – 2 * (ii)
$$-11y+22=0$$
$$y=2$$
Putting back in equation (i).
$$2x-3(2)+12=0$$
$$2x=6\Rightarrow x=-3$$
The center will be (-3, 2)
By the standard equation of circle
$$(x-h{)}^2+(y-k{)}^2=R^2$$
$$(x-(-3){)}^2+(y-2{)}^2=7^2$$
$$(x+3{)}^2+(y-2{)}^2=49$$
$$x^2+9+6x+y^2+4-4y-49=0$$
$$x^2+6x+y^2-4y-36=0$$