Given,lx +my =1 touches the circle x²+y² = a²
Then, length of perpendicular distance from the centre of the given circle i.e. (0,0) on ix +my +1 = 0 is equal to radius
Then, a =\(|\frac { 0-0+1 }{ \sqrt { { l }^{ 2 }+m^{ 2 } } } |\Longrightarrow { a }^{ 2 }=\frac { 1 }{ l^{ 2 }+{ m }^{ 2 } } \Longrightarrow { l }^{ 2 }+{ m }^{ 2 }={ a }^{ -2 }\)

Let the equation of circle be (x - 2)² + (y + 3)² = r²
Intersection point is (5, - 2)
x² + y²- 4x +6y - 10 = 0

\(2b=\frac { 3 }{ 4 } (2a) \Rightarrow b=\frac { 3 }{ 4 } a \Rightarrow e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } \)

\(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 11 } =1\)

Given equation of line is y = mx + 1... (i)
and equation of parabola is y² = 4 x ...(ii)
On putting the value of y from Eq. (i) in Eq. (ii), we get
\((m x+1)^{2}=4 x \Rightarrow m^{2} x^{2}+1+2 m x-4 x=0 \)
\(\Rightarrow x^{2}\left(m^{2}\right)+x(2 m-4)+1=0\)
For the tangent, discriminant is zero i.e. D = 0
\(\therefore (2 m-4)^{2}-4 m^{2} \times 1=0 \quad {\left[\because D=b^{2}-4 a c=0\right]} \)
\(\Rightarrow m=1 \)

Coordinates of center of given circle is (1,2) and it passes through the point(4,6)
Then, radius of the circle is equal to the distance from the centre to a point on a circle
\(\therefore \)  Radius of circle=\(\sqrt { (1-4)^{ 2 }+(2-6)^{ 2 } } =\sqrt { 9+16 } =15\) [by distance formula,distance
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-y_{ 1 })^{ 2 } } \)
Hence, the required equation of the circle is
\((x-1)^{ 2 }+(y-2)^{ 2 }=(5)^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+y^{ 2 }-2x-4y-20=0\)      

Given equation of circle is
x² + y² - 6x + 5y - 8 = 0
(x² - 6x) + (y² + 5y) = 8
(x² - 6x + 3²) + [y² + 5y + \({ \left( \frac { 5 }{ 2 } \right) }^{ 2 }\)] = 3²+\({ \left( \frac { 5 }{ 2 } \right) }^{ 2 }\)+8
(x - 3)²+\({ \left( y+\frac { 5 }{ 2 } \right) }^{ 2 }\)=\({ \frac { 93 }{ 4 } }\)   ...(i)
On computing Eq.(i) with standard form of circlei.e
(x - h)² + (y - k)²= r², we get
h=3, k=-\({ \frac { 5 }{ 2 } }\) and r=\({ \frac { \sqrt { 93 } }{ 2 } }\)

On comparing the given equation with (x - h)² + (y - k)² = r² , we get
\(h=\frac { 1 }{ 2 } ,k=-\frac { 1 }{ 3 } \)
\(and\quad r=\frac { 1 }{ 2 } \)
\((\frac { 1 }{ 2 } ,-\frac { 1 }{ 3 } )\quad and\quad \frac { 1 }{ 2 } \)

Given center is(2,3)
h=2,k=3 and radius (r)=5
On putting these values in equation of circle
\((x-h)^{ 2 }+(y-k)^{ 2 }={ r }^{ 2 }\)   we get 
\((x-2)^{ 2 }+(y-3)^{ 2 }={ 5 }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+4-4x+y^{ 2 }+9-6y=25 \quad [\therefore (A-B)^{ 2 }=A^{ 2 }+B^{ 2 }-2AB]\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-6y+13=25\)
\( \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x+13-25=0\)
\(\therefore { x }^{ 2 }+{ y }^{ 2 }-4x-6x-12=0\)
Which is the required equation of circle

Given equation of ellipse is \(​​\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 16 } =1.\)
\(\text{On comparing with} \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\text{,we get }a=7,b=4\)

Here, a>b, so major axis is along X-axis.​​​​​​​​​​​​​​
\(Vertices,(\pm a,0)=(\pm 7,0)\)

3x² + 5y² = 32

\(Vertices=(0,\pm 20),\quad Major\quad axis=12\quad ,\)
\(Minor\quad axis=20,\quad Eccentricity=\frac { \sqrt { 3 } }{ 2 } \)
\(Foci=(0,\pm 10\sqrt { 3 } ),\quad Latusrectum=10\)

Let F₁ and F₂ be two points where the flag parts are fixed on the ground. The origin O is the mid point of F₁F₂

∴ OF₁=OF₂=\(\frac { 1 }{ 2 } \)F₁F₂=\(\frac { 1 }{ 2 } \) x 8 =4m
∴ Coordinates of F; are (-4, 0) and F₂ are (4, 0)
Let P(α, β) be any point on the track.
∴ PF₁+PF₂ = 0
∴ \(\sqrt { (\alpha +4)^{ 2 }+(\beta -0)^{ 2 } } +\sqrt { (\alpha -4)^{ 2 }+(\beta -0)^{ 2 } } \)
=10
⇒ \(\sqrt { \alpha ^{ 2 }+16+8\alpha +{ \beta }^{ 2 } } \)
=10-\(\sqrt { { \alpha }^{ 2 }+16-8\alpha +{ \beta }^{ 2 } } \)
Squaring both sides, we have
\(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }+8\alpha +16=100+{ \alpha }^{ 2 }+{ \beta }^{ 2 }-8\alpha \)+16
=-20\(\sqrt { { \alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16 } \)
⇒ 16\(\alpha \)-100=-20\(\sqrt { { \alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16 } \)
Squaring both sides again, we have
(16\(\alpha \)-100)2=(-20\(\sqrt { { (\alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16)^{ 2 } } \)
⇒ 256\(\alpha \)²+1000-3200\(\alpha \) = 400(\(\alpha \)²+\(\beta \)²+8\(\alpha \)+16)
⇒ 256\(\alpha \)²+10000-3200\(\alpha \)=400\(\alpha \)²+400\(\beta \)²-3200\(\alpha \)+6400
⇒ 144\(\alpha \)²+400\(\beta \)²=3600
⇒ \(\frac { 144\alpha ^{ 2 } }{ 3600 } +\frac { 400{ \beta }^{ 2 } }{ 3600 } \)=1 ⇒ \(\frac { { \alpha }^{ 2 } }{ 25 } +\frac { { \beta }^{ 2 } }{ 9 } \)=1
Thus required equation of locus of point P is
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } \)=1

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)² + y² = 25.
It is given that the circle passes through point (2, 3).
\(\therefore(2-h)^{2}+3^{2}=25 \)
\(\Rightarrow(2-h)^{2}=25-9 \)
\(\Rightarrow(2-h)^{2}=16 \)
\(\Rightarrow 2-h=\pm \sqrt{16}=\pm 4 \)
\(\text { If } 2-h=4, \text { then } h=-2 . \)
\(\text { If } 2-h=-4, \text { then } h=6 .\)
Equation of required circle is
(x - 6)² + (y - 0)² = (5)²
⇒ x² + 36 - 12x + y² = 25
⇒ x² + y² - 12x + 11 = 0
When h=-2
Equation of required circle is
(x + 2)² + (y - 0)² = (5)²
⇒  x² + 4 + 4x + y² = 25
⇒  x² + y² + 4x - 21 = 0

The equation of the circle is
(x - h)² + (y - k)² = r²... (i)
Since the circle passes through point (4, 1)
∴ (4 - h)² + (1 - k)² = r²
⇒ 16 + h² - 8h + 1 + k² - 2k = r²
⇒ h² + k² - 8h - 2k + 17 = r²...(ii)
Also the circle passes through point (6, 5)
∴ (6 - h)² + (5 - k)² = r²
⇒ 36 + h² - 12h + 25 + k² - 10k = r²
⇒ h² + k² - 12h - 10k + 61 = r²...(iii)
From (ii)and (iii), we have
h²+ k²-8h-2k+ 17=h²+ k²-12h-10k+61
⇒ 4h+ 8k= 44
⇒ h + 2k = 11....(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴  4h + k = 16      ...(v)
Solving (iv) and (v), we have h = 3 and k = 4
Putting value of hand k in (ii), we have
(3)² + (4)² - 8 x 3 - 2 x 4 + 17 = r²
∴  r²=10
Thus equation of required circle is
⇒ (x - 3)² + (y - 4)² = 10
⇒ x² + 9 - 6x + y² + 16 - 8y = 10
⇒ x² + y² - 6x - 8y + 15 =0.