Circles Test 5

Equation of circle touching x-axis at $$(0,0),$$ means centre of circle lie on Y-axis i.e. $$(0,k)$$.

$$(x-0)^{2}+(y-k)^{2}=k^{2}$$

$$S : x^{2}+y^{2}-2ky=0 ...... (1)$$

Circle S touches line $$3x+4y-5=0$$
Perpendicular distance from the centre (0,k) of circle to the given line is equal to the radius of the circle

$$\therefore k=\left | \dfrac{3(0)+4k-5}{\sqrt{3^2+4^2}} \right |=\left|{\dfrac{4k-5}{5}}\right|$$

$$5k=4k-5$$

$$k=-5$$

$$\therefore $$ Equation of circle is

$$x^{2}+y^{2}-(-5)\times2y=0$$

$$\Rightarrow x^{2}+y^{2}+10 y=0$$

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

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 $$x,y$$ )

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

$$x^{2}-y^{2}-4x+4y+3=0$$ 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 $$(x-2)^{2}+(y-2)^{2}=r^{2}$$
$$9+1=r^{2}$$
$$r^{2}=10$$
$$\therefore x^{2}-4x+y^{2}-2y+5=10$$
$$\Rightarrow x^{2}+y^{2}-4x-2y-5=0$$

Given equation of the incirle is $${ x }^{ 2 }+{ y }^{ 2 }+4x-6y+4=0$$

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
$$(x+1)^{2}+(y+2)^{2}=(1)^{2}$$
$$\Rightarrow x^{2}+y^{2}+2x+4y+4=0$$

$$x^{2}+y^{2}-2x+4y+5=0$$
$$(x-1)^{2}+(y+2)^{2} -5+5=0$$
$$\Rightarrow (x-1)^{2}+(y+2)^{2}=0$$
Since, radius is $$0$$, hence its a point

Alternative method:
Here, $$a=b=1$$
$$r=\sqrt{1+4-5}=0$$
Hence, a circle of radius $$0$$. So, its a point.

Given equation is $$x^{2}+y^{2}=2ax$$

Centre $$=(a,0)$$ and radius $$=a$$

Therefore, equation of line having slope $$=$$ $$\dfrac {-1}{2}$$ and passing through $$(4,0)$$
$$y=\dfrac {-1}{2}(x-a)$$
Line $$AB$$: $$2y+x=a$$  -(1)
$$y=\dfrac {a+x}{2}$$
$$d=\left | \dfrac {a}{\sqrt{5}} \right |$$
So, area of $$\triangle AOB =\dfrac {1}{2}\times\ AB\ \times$$ (perpendicular distance of $$Q$$ from $$AB$$ )
$$=\dfrac {1}{2}\times 2a\times \dfrac {a}{\sqrt{5}}$$
$$=\dfrac {a^{2}}{\sqrt{5}}$$