Circles Test 6

$$(x^{2}-a^{2})^{2}+(y^{2}-b^{2})^{2}= 0$$ represents points
$$x= ^{+}_{-}a$$ and $$y= ^{+}_{-}b$$
$$\therefore (a,b), (a,-b), (-a,b)$$ and $$(-a,-b)$$ are four points
These point lie on a circle whose centre is at $$(0,0)$$ and radius is $$\sqrt{a^2+b^2}$$

Let $$(h, k)$$ is a centre of circle , so

The two circles are,

Now, length of chord $$QR=\sqrt{(-4-3)^2+(3-4)^2}=5\sqrt{2}$$

$$ Given-\quad \\ PQ\quad is\quad a\quad chord\quad of\quad a\quad circle\quad with\quad centre\quad O\quad such\quad that\\ OP=OQ=r=PQ\quad when\quad r\quad is\quad the\quad radius\quad of\quad the\quad circle.\\ To\quad find\quad out\quad h=ON,\quad the\quad distance\quad of\quad PQ\quad from\quad O.\\ Solution-\\ \Delta OPQ\quad is\quad equilateral\quad since\quad all\quad its\quad sides=r.\\ Now\quad height\quad of\quad an\quad equilateral\quad triangle\quad with\quad side=a\\ is\quad \frac { \sqrt { 3 }  }{ 2 } a\quad and\quad it\quad bisects\quad the\quad base\quad at\quad right\quad angle.\\ Here\quad a=r.\\ \therefore \quad The\quad height=ON=h=\frac { \sqrt { 3 }  }{ 2 } r.=\quad \\ But\quad ON\quad is\quad the\quad distance\quad of\quad the\quad chord\quad PQ\quad from\quad O\quad since\\ the\quad perpedicular\quad distance\quad of\quad a\quad chord\quad from\quad the\quad centre\quad \\ bisects\quad the\quad chord\quad at\quad right\quad angle.\\ So\quad h=ON=\frac { \sqrt { 3 }  }{ 2 }r\quad is\quad the\quad distance\quad of\quad the\quad chord\quad PQ\quad from\quad O.\\ Ans-\quad Option\quad D.\\ \\ \quad \\ \quad \quad \\ \\  $$

Since, the lines $$\displaystyle 2x - 3y = 5 \dots (1)$$ and $$\displaystyle 3x - 4y = 7 \dots (2)$$ passes through the center of the circle.

Given, median of the equilateral triangle is $$3a$$ say $$LD $$
In $$\displaystyle \Delta LMD$$, we have

$$x^{2}-y^{2}-2x+4y-3=0$$

Given diameter of circle is $$x+y=1$$
Diameter crosses the x-axis at $$(1,0)$$ and y-axis at $$(0,1)$$
Distance between these two points i.e. $$d=\sqrt{2}$$
$$\Rightarrow \displaystyle r=\frac{\sqrt{2}}{2}$$
Center of the circle is $$\displaystyle (\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } )$$
Equation of circle is 
$$\displaystyle { (x-\frac { 1 }{ 2 } ) }^{ 2 }+{ (y-\frac { 1 }{ 2 } ) }^{ 2 }=\frac { 1 }{ 2 } $$
$$x^{2}+y^{2}-x-y=0$$

Given lines $$2x+3y+1= 0$$ and $$3x-y-4= 0$$
Point of intersection of these lines is the center of circle.
So, center is at (1,-1).

Also, given circumference $$=10\pi$$
$$\Rightarrow 2\pi r=10\pi $$
$$\Rightarrow r=5$$

Hence the equation of circle is
$$(x-1)^2 +(y+1)^2=25$$
$$x^2+y^2-2x+2y-23=0$$