CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 3 - MCQExams.com

Any point on the line$$ x+ y= 1$$ can betaken as$$(t,1-t)$$

Equation of chord with this as midpoint is

$${y(1-t)-2a(x+t) = (1-t)^2-4at}$$

It passes through $$(a,2a)$$

therefore, $${t^2-2t+2a^2-2a+1=0}$$

This should have two distinct real roots so

$$-a<0$$

$$0<a<1$$

length of latus rectum $$< 4$$

so answer is $$1,2$$

$$\textbf{Step -1: Find the radius of the circle.}$$

                $$\text{As we know that, the standard equation of a circle is:}$$

                $$(x-h)^2+(y-k)^2=r^2\text{ where }(h,k)\text{ is the centre and }r\text{ is the radius.}$$

                $$\text{Here, the equation of the circle will be,}$$

                $$(x-2)^2+(y+1)^2=r^2\ldots(i)$$

                $$\text{and the point }(3,6)\text{ will satisfy this equation.}$$

                $$\therefore (3-2)^2+(6+1)^2=r^2$$

                $$\Rightarrow 1+49=r^2$$

                $$\Rightarrow r^2=50$$

$$\textbf{Step -2: Find the equation of the required circle.}$$

                $$\text{On putting the value of }r^2\text{ in equation }(i),\text{ we get}$$

                $$(x-2)^2+(y+1)^2=50$$

                $$\Rightarrow x^2+4-4x+y^2+1+2y=50$$

                $$\Rightarrow x^2+y^2-4x+2y=45$$

$$\textbf{Hence, option A is correct.}$$

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

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

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

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 $$ = \sqrt {5} $$

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

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

$$(x – 4) ^2  + (y)^2 = 49$$ 

$$(x - 3) ^2  + (y+ 4)^2 = 4$$

$$\implies center  = (3,-4) , r = 2$$

Since $$C = (3,-4)$$

The equation of a circle of radius r and centre the origin is

$$x^2+y^2=r^2$$

Here the center is (a, b)

so Radius , $$r = \sqrt{a^2+b^2}$$