Conic Sections Test - 20

Equation of any circle passing through the point of intersection of 

Consider Equation, $$\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$$ to represent an ellipse equation.
$$a>0,b>0,a\neq b$$
Given,equation $$\displaystyle\frac{x^2}{(8-t)}+\displaystyle\frac{y^2}{(t-4)}=1$$
$$\Rightarrow (8-t)>0\;$$ and $$\;(t-4)>0,(8-t)\neq(t-4)$$
$$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$$ {$$t\neq6$$}
$$\Rightarrow t\in(4,8)-$${$$6$$}

We know the normal of a circle passes through it's center.
So the given points $$(3,4)$$ and $$(1,2)$$ are end point of the diameter.
and  the centre of the circle is mid point of the given points, $$\displaystyle \equiv \left( 2,3 \right) $$
and Radius $$=\displaystyle \frac { \sqrt { 4+4 }  }{ 2 } =\frac { 2\sqrt{ 2 }  }{ 2 } =\sqrt { 2 } $$
Hence required circle is, $$\displaystyle (x-2{ ) }^{ 2 }+(y-3{ ) }^{ 2 }=2$$
$$\Rightarrow \displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x-6y+11=0$$

Intersection of any 2 diameters gives us the center

$$2x – 3y = 5 -eq.1$$

$$3x – 4y = 7 -eq.2$$

$$3 \times eq.1 – 2 \times eq.2$$

$$\implies -9y + 8y = 15 - 14$$

$$\implies y = -1 \implies x = 1$$

Center $$= (1,-1)$$

And given $$A = 154$$

$$\implies \pi r^2 = 154$$

$$r^2 = \dfrac{154}{22} \times 7 = 49$$

$$r = 7$$

Equation of circle is

$$(x - 1) ^2 + (y+1)^2 = 7^2$$

$$\implies x^2 + y^2 -2x + 2y = 47$$

The given equation can be written as $$(x - 1) ^2 + (y - 1)^2 = 1^2$$

Any point on this circle is given by $$(1 + \cos \theta , 1 + \sin \theta)$$

$$4x + 3y = 7 + 4 \cos \theta + 3 \sin \theta = f(\theta)$$

If $$f = a \cos \theta + b \sin \theta + c$$

$$max = c + \sqrt{a^2 + b^2} , min = c - \sqrt{a^2 + b^2}$$

$$\implies max \, f(\theta) + min \, f(\theta) = 2c =2 \times 7 = 14$$

Intersection of any 2 diameters gives us the center

$$2x – 3y = 5 -eq.1$$

$$3x – 4y = 7 -eq.2$$

$$3 \times eq.1 – 2 \times eq.2$$

$$\implies -9y + 8y = 15 - 14$$

$$\implies y = -1 \implies x = 1$$

Center = (1,-1)

And given A = 154

$$\implies \pi r^2 = 154$$

$$r^2 = \dfrac{154}{22} \times 7 = 49$$

$$r = 7$$

Equation of circle is

$$(x - 1) ^2 + (y+1)^2 = 7^2$$

$$\implies x^2 + y^2 -2x + 2y – 47 = 0$$

Radius of the circle $$ = $$ dist. between the center and given pt. on the circle.

Distance between two points $$

\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }

\right) $$ can be calculated using the formula 

$$ \sqrt { \left( { x }_{ 2 }-{

x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$

Distance between the points $$ (x-2,x+1) $$ and D $$ (4,4) $$ 

$$= \sqrt { \left( 4-x + 2 \right) ^{ 2 }+\left( 4 - x - 1 \right) ^{ 2 } } $$ 

$$= \sqrt { \left( 6-x \right) ^{ 2 }+\left( 3 - x \right) ^{ 2 } } $$ 

$$= \sqrt { 36 + {x}^{2} - 12x + 9 + {x}^{2} - 6x } = \sqrt { 2{x}^{2} -18x + 45 } $$

Given, diameter $$ = 2 \sqrt {5} $$ $$\Rightarrow$$ Radius $$ = \sqrt {5} $$

$$ \Rightarrow \sqrt { 2{x}^{2} -18x + 45 }

= \sqrt {5} $$
Squaring both sides,
$$ 2{x}^{2} -18x + 45 = 5 $$
$$ 2{x}^{2} -18x + 40 = 0 $$
$$ {x}^{2} -9x + 20 = 0 $$
$$ (x-5)(x-4) = 0 $$
$$ x = 4 $$ or $$ 5 $$

The equation of circle is $$x^2+y^2=r^2$$

According to the question

Lets suppose that,

X and Y are two circle touch each other at P.

AB is the common tangent to circle X and Y at point A and B.

 According In the given figer,

In triangle $$PAC,\ \angle CAP=\angle APC=\alpha $$

Similarly $$CB=CP,\ \angle CPB=\angle PBC=\beta $$

Now triangle APB,

  $$ \angle PAB+\angle PBA+\angle APB=180 $$

 $$ \alpha +\beta +\left( \alpha +\beta  \right)=180 $$

 $$ 2\alpha +2\beta =180 $$

 $$ \alpha +\beta =90 $$

 $$ \therefore \ \angle APB=90=\alpha +\beta . $$

This is the required solution.

If $$(-g,-f)$$ is the center and $$r$$ is radius

The $$(x + g) ^2  + (y+ f)^2 = r^2     $$ is the equation of the circle

There $$C = (4,0) , r = 7$$

$$\implies (x - 4) ^2  + (y - 0)^2 = 7^2$$

$$(x – 4) ^2  + (y)^2 = 49$$