Circles

Given four points \((2 k, 3 k),(1,0),(0,1),(0,0)\)

since \((1,0)\) and \((0,1)\) subtends an angle of \(90^{\circ}\) at \((0,0)\)
Circle passes through \((1,0),(0,1),(0,0)\)

\((x-1)(x-0)+(y-0)(y-1)=0\)
\( x^2+y^2-x-y=0 \)

\((2 k, 3 k)\) lies on circle.
\( \Rightarrow 4 k^2+9 k^2-2 k-3 k=0 \)
\( k=\frac{5}{13}\)

Given four points \((2 k, 3 k),(1,0),(0,1),(0,0)\)

since \((1,0)\) and \((0,1)\) subtends an angle of \(90^{\circ}\) at \((0,0)\)
Circle passes through \((1,0),(0,1),(0,0)\)

\((x-1)(x-0)+(y-0)(y-1)=0\)
\( x^2+y^2-x-y=0 \)

\((2 k, 3 k)\) lies on circle.
\( \Rightarrow 4 k^2+9 k^2-2 k-3 k=0 \)
\( k=\frac{5}{13}\)

\( x+(k-1) y+3=0 \)
\( 2 x+k^2 y-4=0\)

Since these \(2\) lines are perpendicular

\( \frac{1}{1-k}\left(-\frac{2}{k^2}\right)=-1 \)
\( 2=k^2(1-k) \)
\( \Rightarrow k^3-k^2+2=0\)
\(k^3-k^2+2=(k+1)\left(k^2-2 k+2\right)=0\),

and since \(k\) is real, \(k=-1\)

$\therefore$ Lines are :

\(x-2 y+3=0 \) and \(2 x+y-4=0\)

then \( x=1\) and \(y=2\)
$\therefore$ Centre is \((1,2)\)
Circle will be \(x^2+y^2-2 x-4 y=0\)
Line \(x-y+2=0\) will intersect at \(A(-1,1)\) and \(B(2,4)\)

\( \therefore(A B)^2=(2+1)^2+(4-1)^2 \)
\( =9+9=18\)

The circle \(C_1\) has a diameter of \(10 \), so its radius is \(R=5\). It passes through the origin \((0,0)\) and is contained in the closed half-plane \(x \geq 0\).

Therefore centre is \((5,0)\)

The equation of \(C_1\) is:

\((x-5)^2+y^2=25\)

\(\Rightarrow x^2+y^2-10 x=0\)

The line \(y=x\) is a chord of \(C_1\). We find the intersection points of \(y=x\) and \(C_1\) :

\((x-5)^2+x^2=25 \)

\( \Rightarrow x^2-10 x+25+x^2=25 \)

\( \Rightarrow 2 x^2-10 x=0 \)

\( \Rightarrow 2 x(x-5)=0\)

The intersection points are \((0,0)\) and \((5,5)\). These are the endpoints of the diameter of circle \(C_2\).

The center \(M\) of \(C_2\) is the midpoint of the chord:

\(M\left(\frac{5}{2}, \frac{5}{2}\right)\)

Given point is \(P(2,3)\)

In any circle, the chord passing through a point \(P\) that is farthest from the center is the one perpendicular to the radius (or segment) \(M P\).

Now the slope of the segment \(M P\) is:

\(m_{M P}=\frac{3-2.5}{2-2.5}=\frac{0.5}{-0.5}=-1\)

The slope \(m\) of the required chord is

\(m=-\frac{1}{-1}=1\)

Now equation of required chord is \(y-3=1(x-2)\)

\(\Rightarrow y-3=x-2 \Rightarrow x-y+1=0\)

we get \( a=-1\) and \( b=1\)

Now \(a-b=-1-1=-2\)

$$\begin{gathered} \left|r_1-r_2\right|< C_1{C}_2<r_1+r_2 \\ Let C_1(2,2) \& r_1=2 \\ {C}_2(2,6) \\ Now \left|r_1-r_2\right|< C_1{C}_2<r_1+r_2 \\ \left|r_2-2\right|< 4\sqrt{2}<2+r_2 \\ From above \\ {r}_2> 4\sqrt{2}-2 \\ \& \left|r_2-2\right|< 4\sqrt{2} \\ -4\sqrt{2}<r_2-2<4\sqrt{2} \\ {r}_2<4\sqrt{2}+2 \\ {r}_2 [4\sqrt{2}-2 , 4\sqrt{2}+2] \\ a=[4\sqrt{2}-2 , b=4\sqrt{2}+2 \\ a+b=4\sqrt{2}-2+4\sqrt{2}+2 \\ a+b=8\sqrt{2} \end{gathered}$$

$$\begin{gathered} \left|r_1-r_2\right|< C_1{C}_2<r_1+r_2 \\ Let C_1(2,2) \& r_1=2 \\ {C}_2(2,6) \\ Now \left|r_1-r_2\right|< C_1{C}_2<r_1+r_2 \\ \left|r_2-2\right|< 4\sqrt{2}<2+r_2 \\ From above \\ {r}_2> 4\sqrt{2}-2 \\ \& \left|r_2-2\right|< 4\sqrt{2} \\ -4\sqrt{2}<r_2-2<4\sqrt{2} \\ {r}_2<4\sqrt{2}+2 \\ {r}_2 [4\sqrt{2}-2 , 4\sqrt{2}+2] \\ a=[4\sqrt{2}-2 , b=4\sqrt{2}+2 \\ a+b=4\sqrt{2}-2+4\sqrt{2}+2 \\ a+b=8\sqrt{2} \end{gathered}$$

By family of curve equation of circle will be

$\Rightarrow S_1+\lambda S_2=0$

$\Rightarrow (x^2+2y^2-6x-12y+23)$$+\lambda (4x^2+2y^2-20x-12y+35)=0$

⇒ for circle coeff. of $x^2=$ coeff. of $y^2$

^{$\Rightarrow \lambda =\frac{1}{2}$}

So equation of circle is

$\Rightarrow x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0$

Centre $\left(\frac{8}{3},3\right):$ Radius $r=\sqrt{\frac{47}{18}}$

$\therefore ab+18r^2=8+47=55$

Let \(A(4,6), B(-1,5)\) and \(C(0,0)\)
\(m_{A B}=\frac{1}{5}\) and \(m_{B C}=-5\)
\(\Rightarrow A C\) is diameter

Equation of circle is

\((x-4)(x-0)+(y-6)(y-0)=0\)

\(x^2+y^2-4 x-6 y=0\)

Point \((k,3k)\) lies on the circle, therefore

\(10 k^2-22 k=0\)

\(k=\frac{11}{5}\)

and radius \(r=\frac{1}{2} \sqrt{52}\)

Now \(10 k+r^2=10 \times \frac{11}{5}+\frac{1}{4} \times 52=35\)

$$\begin{gathered} Let A=\left(4,6\right), B=(-1,5) and C=\left(0,0\right) \\ {m}_{AB}=\frac{1}{5} and m_{BC}=-5 \\ \Rightarrow AC is diameter \\ Eq of circle is x^2+y^2-4x-6y=0 \\ \Rightarrow 10k+r^2=35 \end{gathered}$$

$$\begin{gathered} x^2+y^2-\alpha x-\beta y+\gamma =0 \\ g= \frac{-\alpha }{2}, f= \frac{-\beta }{2} \\ centre \left(\frac{\alpha }{2},\frac{\beta }{2}\right) \\ Now a=\frac{\alpha }{2} \Rightarrow \alpha =2a \\ \& b =2\sqrt{\frac{{\beta }^2}{4}-\gamma } \Rightarrow b^2={\beta }^2-4\gamma \\ Hence, (2a, b^2)=(\alpha , {\beta }^2-4\gamma ) \end{gathered}$$

Given \(x^2+y^2=4\)

Put \(y=0\) we get \(x=\pm 4\)

we get \(A(2,0), B(-2,0)\)

Let point of intersection $R (h, k)$

$m_{BR}=m_{BP}\Rightarrow \frac{k}{h+2}=\frac{2\sin \alpha }{2\cos \alpha +2}$

$\Rightarrow \frac{k}{h+2}=\tan \frac{\alpha }{2}$

$m_{AR}=m_{AQ}\Rightarrow \frac{k}{h-2}=\frac{2\sin \beta }{2\cos \beta -2}=\frac{\sin \beta }{\cos \beta -1}$

$=-\cot \frac{\beta }{2}$

Now, $\frac{\alpha }{2}-\frac{\beta }{2}=\frac{\pi }{4}$

$\Rightarrow \tan \left(\frac{\alpha }{2}-\frac{\beta }{2}\right)=\tan \frac{\pi }{4}=1$

$\Rightarrow \frac{\tan \frac{\alpha }{2}-\tan \frac{\beta }{2}}{1+\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=1$

$$\begin{gathered} \Rightarrow \frac{\frac{k}{h+2}+\frac{h-2}{k}}{1+\left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right)}=1 \\ \Rightarrow \frac{k^2+h^2-4}{k(h+2)+k\left(2-h\right)}=1 \end{gathered}$$

$\Rightarrow \frac{h^2+k^2-4}{4k}=1$

Hence locus is

$x^2+y^2–4y–4=0$

The circle $C_1$ has a radius of 2 and touches

both coordinate axes in the second quadrant.

Therefore, its center is (-2, 2).

The equation of circle $C_1$ is

${\left(x+2\right)}^2+{\left(y-2\right)}^2=2^2$

Let the circle $C_2$ with center (2, 5) and radius r.

${\left(x-2\right)}^2+{\left(y-5\right)}^2=r^2$

The distance between the centers of

circles $C_1$and $C_2$is:
$d=\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(5-2\right)}^2}=5$

For the circles to intersect at exactly two points,

the following condition must hold:

$|r - 2| < 5 < r + 2$

$|r - 2| < 5\Rightarrow -3<r<7$

$5 < r + 2 \Rightarrow r > 3$

On combining, we get $3 < r < 7.$

$$\begin{gathered} \text{From the inequality} 3 < r < 7 \\ \text{we can identify} \alpha = 3 \text{and} \beta = 7. \end{gathered}$$

$\Rightarrow 3\beta - 2\alpha = 3\left(7\right) - 2\left(3\right) = 15$

$$\begin{gathered} x^2+y^2-\alpha x-\beta y+\gamma =0 \\ g= \frac{-\alpha }{2}, f= \frac{-\beta }{2} \\ centre \left(\frac{\alpha }{2},\frac{\beta }{2}\right) \\ Now a=\frac{\alpha }{2} \Rightarrow \alpha =2a \\ \& b =2\sqrt{\frac{{\beta }^2}{4}-\gamma } \Rightarrow b^2={\beta }^2-4\gamma \\ Hence, (2a, b^2)=(\alpha , {\beta }^2-4\gamma ) \end{gathered}$$