Circles Test 25

$${\textbf{Step 1: Compare the coordinates of given points with the end points of line.}}$$

                $${\text{Given end point of diameter are}}$$$$\left( { - 6,3} \right)$$$${\text{and}}$$$$\left( {6,4} \right)$$

                $${\text{Comparing with}}$$ $$\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$$

                $${\text{Here,}}$$$${x_1} =  - 6,{y_1} = 3,{x_2} = 6,{y_2} = 3$$

$${\textbf{Step 2: Put these values in eqn of mid-point of a line.}}$$

                $${\text{Now, Centre of a circle is mid - point of the diameter}}{\text{.}}$$

                $${\text{Mid-point of a line is given by=}}$$ $$\left( {\frac{{{x_1} + {x_2}}}{2}} \right),\left( {\frac{{{y_1} + {y_2}}}{2}} \right)$$

                $${\text{Therefore, centre of circle is   =}}$$$$\left( {\frac{{ - 6 + 6}}{2}} \right),\left( {\frac{{3 + 4}}{2}} \right)$$

                                                                      $$ = \left( {0,\frac{7}{2}} \right)$$

$${\textbf{Final Answer: Hence, the required answer is}}$$ $$\mathbf{\left( {0,\frac{7}{2}} \right).}$$

Let the equation of the general form of the required circle be 

$$x^2+y^2+2gx+2fy+c=0$$................(1)

According to the problem, the above equation of the circle passes through the points $$(0, 6), (0, 0)$$ and $$(8, 0)$$. Therefore,

$$36 + 12f + c = 0$$  ………. (2)

$$ c = 0$$                       ……………. (3)

$$64 + 16g + c = 0$$  ……………. (4)

Putting $$c = 0$$ in (2), we obtain $$f = -3$$. Similarly put $$c = 0$$ in (4), we obtain $$g=-4$$

Substituting the values of $$g, f$$ and $$c$$ in (1), we obtain the equation of the required circle as:

$$x^2+y^2+2(-4)x+2(-2)y+0=0$$ that is 

$$x^2+y^2-8x-4y+0=0$$ can be rewritten as

$$x^2+y^2-8x-4y+16+9=0+16+9$$

$$(x-4)^2+(y-3)^2=25$$

Therefore, the equation of circle is $$(x-4)^2+(y-3)^2=25$$.

Equation of circle having centre of origin $$(0,0)$$ and radius $$=r$$,

$$S_{1};x^{2}+y^{2}=r^{2}----(1)$$

$$\therefore f(m_{1},\dfrac{1}{m_{2}}),f(m_{2},\dfrac{1}{m_{2}}),--f(m_{4},\dfrac{1}{m_{4}})$$
These points lie on $$S_{1}$$.

Let $$\displaystyle f(m, \frac{1}{m})$$ is point lie on S_{1},

$$m^{2}+\dfrac{1}{m^{2}}=r^{2}$$

$$m^{4}+1-r^{2}m^{2}=0$$

$$m^{4}-1-r^{2}m^{2}+1=0$$

$$m_{1},m_{2},m_{3}$$ and $$m_{4}$$ are roots of this equation

So, $$m_{1}m_{2}m_{3}m_{4} =1$$

Any point on the lines $$7x + y + 3 = 0$$ is $$Q (t, -3 -7 t),       t  \in R.$$
Now P $$(h, k)$$ is image of point Q in the line $$x - y +1 = 0$$
Then, $$\displaystyle \frac{h - t}{1} = \frac{k - (-3 - 7 t)}{-1}$$
         $$ = \displaystyle - \frac{2 (t - (-3 - 7t) + 1)}{1+1}$$
         $$ = -8t - 4$$
$$\Rightarrow       (h, k) \equiv (-7t - 4, t + 1)$$
This point lies on the circle $$x^2 + y^2 = 9$$
$$\Rightarrow (-7t - 4)^2 + (t + 1)^2 = 9$$
$$\Rightarrow 50 r^2 + 58t + 8 = 0$$
$$\Rightarrow 25r^2 + 19t + 4 = 0$$
$$ \Rightarrow (25 t + 4) (t + 1) = 0$$
$$\Rightarrow t = -4/25, t = 1$$
$$\Rightarrow (h, k) = \displaystyle \left ( -\frac{72}{25}, \frac{21}{25} \right ) $$ or $$(3, 0)$$

writing in intercept form, we get 
$$\dfrac{x}{8}+\dfrac{y}{6}=1$$ and 

$$\dfrac{x}{6}+\dfrac{y}{8}=1$$
Hence the point are $$(0,8),(0,6)$$ and $$(6,0)(8,0)$$
Hence
The circle cuts the x axis at $$(6,0)(8,0)$$
Hence x-coordinate of the center will be 
$$=6+\dfrac{8-6}{2}$$
$$=7$$
The circle cuts the x axis at $$(0,6)(0,8)$$
Hence y-coordinate of the center will be 
$$=6+\dfrac{8-6}{2}$$
$$=7$$
The center of the circle will lie at 
$$(7,7)=C$$
Now let the point $$(6,0)$$ be A.
Hence 
R=radius
$$CA$$$$=\sqrt{50}$$
Hence, the equation of the circle will be 
$$(x-7)^{2}+(y-7)^{2}=50$$
$$x^{2}+y^{2}-14x-14y+98=50$$
Or 
$$x^{2}+y^{2}-14x-14y+48=0$$

Let $$s: x^2 + y^2 + 2gx + 2fy +c = 0$$ be the equation of the circle

$$S(0,2) = 0$$

$$\implies 4 + 4f + c =0 -eq.1$$

Intercept on x-axis is given by

$$2\sqrt{g^2 - c} = 4$$

$$g^2 – c = 4 -eq.2$$

Since x = 0 is a tangent

Radius = perpendicular distance from the center to tangent

$$\sqrt {g^2 + f^2 - c} = \dfrac{|-g|}{\sqrt{1 + 0}$$

$$ g^2 + f^2 - c  = g^2 $$

$$f^2 = c$$

Substituting in eq.1

$$4 + 4f + f^2 =0 \implies (f+2)^2 = 0 \implies f = -2 $$

$$\implies c = 4$$

From eq.2

$$g^2 – 4 = 4$$

$$g = \pm 2 \sqrt{2}$$

Equation of circle is

$$x^2 +y^2 \pm 4\sqrt 2 x – 4y + 4 = 0$$

$$\implies x^2 +y^2 – 4(\sqrt 2 x + y) + 4 = 0$$