Conic Sections Test - 14

$$AP^{2}+AB^{2}+PC^{2}=C$$
$$(x-x_{1})^{2}+(y-y_{1})^{2}+(x-x_{2})^{2}-(y-y_{2})^{2}+(x-x_{3})^{2}+(y-y_{3})^{2}=c$$
$$3x^{2}-(2x_{1}+2x_{2}+2x_{3})x+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+3y^{2}-2(y_{1}+y_{2}+y_{3})+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=c$$
$$x^{2}+y^{2}-\frac{2}{3}(x_{1}+x_{2}+x_{3})x-\frac{2}{3}(y_{1}+y_{2}+y_{3})c+(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-c)=0$$
So, center of this circle is $$(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3}),$$
centroid of triangle is a center of that circle.

$$2x-3y=5----(1)$$
$$3x-4y=7----(2)$$
are diameter of circle so, point of intersection is centre of circle.
from (1) & (2), $$x=1, y=-1$$
$$\therefore $$ area of circle is $$154$$
$$=\pi r^{2}=154$$
$$r^{2}=\dfrac{7}{22}\times154$$
$$=\dfrac{7\times14}{2}$$
$$r^{2}=49$$
$$r=7$$
$$\therefore$$ Equation of circle with centre $$(1,-1)$$ and radius $$7$$ units
$$(x-1)^{2}+(y+1)^{2}=7^{2}$$
$$\Rightarrow x^{2}+y^{2}-2x+2y-47=0$$

Given that:

$$2x+3y+1=0$$  and  $$3x-y-4=0$$ lie along diameter of circle.

$$2x-3y=5----(1)$$
$$3x-4y=7----(2)$$
Intersection of this lines given centre of circle
$$\therefore$$ from (1) & (2),
$$n=1, y=-1$$
So, equation of circle is
$$(x-1)^{2}+(y+1)^{2}=(7)^{2}$$
$$\Rightarrow x^{2}+y^{2}-2x+2y-47=0$$

The image of circle w.r.t same line means image of centre w.r.t that line without changing radius
$$x^{2}+y^{2}-6x-4y+12=0$$
Centre $$=(3,2)$$ Radius$$=1$$
Image of $$(3,2)$$ w.r.t $$x+y-1=0$$
$$\dfrac {x-3}{1}=\dfrac {y-2}{1}=-2\dfrac {(3+2-1)}{1^{1}+1^{2}}=-4$$
$$x=-1$$, $$y=-2$$
Then equation of image of circle is
$$(x+1)^{2}+(y+2)^{2}=(1)^{2}$$
$$\Rightarrow x^{2}+y^{2}+2x+4y+4=0$$

Coordinates of the centre of the circle $$\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 1,2 \right) $$ 

Let the equation of line be 
$$\displaystyle \frac{x}{a}+\frac{y}{b}=1$$
where $$a$$ and $$b$$ are the x-intercept and y-intercept.
Then the coordinates of A and B are $$(a,0)$$ and $$(0,b)$$
Distance of origin to the line is $$c$$
$$\Rightarrow c=\displaystyle \frac{|-1|}{\sqrt{(\dfrac{1}{a})^2+(\dfrac{1}{b})^2}}$$
$$\Rightarrow \displaystyle \frac{1}{c}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$$
$$\Rightarrow \displaystyle \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$$   ....(1)
Let the center of circle through O, A, B be $$(h,k)$$
Points $$O(0,0), A(a,0),(0,b)$$ forms a right triangle.
So, the center of circle is the mid-point of AB i.e.$$\displaystyle(\frac{a}{2},\frac{b}{2})$$
$$\Rightarrow h=\displaystyle \frac{a}{2}, k=\frac{b}{2}$$
$$\Rightarrow a=2h, b=2k$$
Substitute this value in (1), we get
$$ \displaystyle \frac{1}{c^2}=\frac{1}{4h^2}+\frac{1}{4k^2}$$
$$\Rightarrow \displaystyle \frac{4}{c^2}=\frac{1}{h^2}+\frac{1}{k^2}$$
$$\Rightarrow 4c^{-2}=x^{-2}+y^{-2}$$ (Replacing $$h,k$$ by $$x,y$$ )

$$x^{2}-y^{2}-4x+4y+3=0$$ are pair of lines along the diameters  of required circle.
So, point of intersection this pair of straight line is centre of required circle.
For point of intersection
Partial differential w.r.t $$x$$; $$2x-4=0$$ --------(1)
Partial diff. equation  w rt $$y$$; $$-2y+2=0$$-------(2)
from (1) & (2), $$x=2 y=1$$
 $$\therefore$$ Equation of circle,
$$(x-2)^{2}+(y-1)^{2}=r^{2}$$.
So, (-1,2) lies on $$(x-2)^{2}+(y-2)^{2}=r^{2}$$
$$9+1=r^{2}$$
$$r^{2}=10$$
$$\therefore x^{2}-4x+y^{2}-2y+5=10$$
$$\Rightarrow x^{2}+y^{2}-4x-2y-5=0$$

This  is  of  the  form  $$SP + SP^{'} = 2a$$
$$\therefore PP^{'} = 2ae$$
$$ =\sqrt{(2+2)^{2}}$$
$$ = 4 $$
$$2ae = 4$$
But $$ 2a = 8$$ (from the equation given)
$$\therefore e= \dfrac{1}{2}$$