Explanation
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
\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)
As the vertex is(1,1) and focus is (3,1), whose ordinate is same its axis of symmetry is y = 1.
And as vertex is equidistant from foci and directrix, and latter is perpendicular to axis of symmetry.
Directrix is x =1
As parabola is the locus of a point whose distance from directrix x+1 = 0 and focus (3,1)
Its equation is (x-3)^2+(y-1)^2= (x+1)^2
\Rightarrow x^2-6X+9+y^2-2Y+1 = x^2+2X+1
\Rightarrow y^2-2y+9 = 8x
\Rightarrow (y-1)^2 = 8(x-1)
So, option C is the answer.
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}
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