The equation of given circule is x2+y2−2x−4y−20=0
whose centere is(1,2) and radius is 5.
Now this circles touches another circle externally at point (3,7).
Let (a,b) be the center of required circle then(3,7) is the mid point of line segment joining the centres of two circles.
\(\therefore \frac { a+1 }{ 2 } =3\quad and\quad \frac { b+2 }{ 2 } =7\)
\(\therefore \quad a=5\quad and\quad b=12\)
thus equation of required circle is
\(\left( { x-5 } \right) ^{ 2 }+\left( y-12 \right) ^{ 2 }=\left( 5 \right) ^{ 2 }\)

x²+ y² - 6x - 6y + 9 = 0; x²+ y² - 30x - 30y + 225=0

\(\left( \frac { 5 }{ 4 } ,-\frac { 5 }{ 8 } \right) ;\frac { 3\sqrt { 5 } }{ 8 } \).

(1,2); \(\sqrt { \frac { 27 }{ 2 } } \).

Equation of parabola, whose axis is X-axis and passing through the point (2,3), will be of the formy y² = 4ax
On putting x = 2, y = 3, we get
9 = -4a (2) \(\Rightarrow a=\frac { 9 }{ 8 } \)

Parabola is symmetrical about Y-axis and passes through (2, -3)
So, equation of parabola is of the form x² = -4 day
On putting x= 2, y=-3, we get
4 = -4a(-3)\(\Rightarrow a=\frac { 1 }{ 3 } \)

Given, center (h,k)=(2,0)and circle touches Y-axis.
Radius(r)=x-coordinate of center=2 so, the equation of circle is
(x - 2)² + (y - 0)² = 2²  [\(\because\)(x - h)² + (y - k)² = r²]
x² + 4 - 4x + y² = 4     [\(\because\)(A - B)² = A² + B² - 2AB]
x² + y² - 4x + 4 = 4
x² + y² - 4x = 0
which is the required equation of circle

Given, centre (h, k) = (1, 2)
and circle touches on X-axis.
\(\therefore\) Radius (r) = y-coordinate of centre = 2
So, equation of circle is
\((x-1)^{2}+(y-2)^{2}=2^{2} \quad\left[\because(x-h)^{2}+(y-k)^{2}=r^{2}\right] \)
\(\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4=4 \)
\(\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \)
\(\Rightarrow x^{2}+y^{2}-2 x-4 y+1=0\)
which is the required equation of circle.

The equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
\(\therefore \ { a }^{ 2 }=25\ and \ { b }^{ 2 }=9\)
\(\text { Eccentricity of ellipse }=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\text { So the co-ordinates of foci are } (\pm 4,0)\)
\(\text { Let equation of required hyperbola is}\)
\(\frac { { x }^{ 2 } }{ { a' }^{ 2 } } -\frac { y^{ 2 } }{ { b }'^{ 2 } } =1\)
\(\text { Let e' be the eccentricity of required hyperbola then,}\)
\(a'e'=4\Rightarrow 2e'=4\Rightarrow e'=2\)
\(\therefore b'^{ 2 }={ a' }^{ 2 }({ e' }^{ 2 }-1)\Rightarrow b'^{ 2 }=4(4-1)=12\)
\(\text { Thus equation of required hyperbola is}\)
\(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

We know that, latusrectum ia a line perpendicular to the axis and passing through focus whose length is 4a.

Given, \({ x }^{ 2 }=12y\), which is of the form \({ x }^{ 2 }=4ay\)
\(4a=12=Length\quad of\quad ACB\)
Focus C=(0,3)
\(Area\quad of\quad \triangle OAB=\frac { 1 }{ 2 } \times AB\times OC=\frac { 1 }{ 2 } \times 12\times 3\)
\(=6\times 3=18sq\quad units\)

\(\frac { { y }^{ 2 } }{ \frac { 36 }{ 5 } } -\frac { { x }^{ 2 } }{ 4 } -=1\)
Here , \({ a }^{ 2 }=\frac { 36 }{ 5 } \)  and b² = 4  \(c=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \)
\(=\sqrt { \frac { 36 }{ 5 } +4 } =\sqrt { \frac { 56 }{ 5 } } =2\sqrt { \frac { 14 }{ 5 } } \)
\(\left( 0,\pm \frac { 2\sqrt { 14 } }{ \sqrt { 5 } } \right) ,\left( 0,\pm \frac { 6 }{ \sqrt { 5 } } \right) ,\frac { \sqrt { 14 } }{ 3 } ,\frac { 4\sqrt { 5 } }{ 3 } \)

\(\sqrt { { (x-1) }^{ 2 }+{ (y+1) }^{ 2 } } =\frac { 1 }{ 2 } .\frac { \left| x-y-3 \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \)
\(\Rightarrow 8[({ x }^{ 2 }+1-2x)+({ y }^{ 2 }+1+2y)]\)
\(={ x }^{ 2 }+{ y }^{ 2 }+9-2xy+6y-6x\)
Ans. \(7{ x }^{ 2 }+7{ y }^{ 2 }+2xy+10x-10y+7=0\)

Since the axes of the parabola is a line perpendicular to the directrix and passing through the vertex (2,1).

So the equation of lines is x+y+λ=0.
∴ 2+1+λ=0⇒λ=−3
∴ equation of directrix is x+y−3=0....(i)
equation of directrix is x−y+1=0....(ii)
Solving (i) and (ii),we have
x=1 and y=2
So the co−ordinate of B are(1,2)
Now A is mid point of  BS
\( \therefore \frac { { x }_{ 1 }+1 }{ 2 } =2\quad and\frac { { y }_{ 1 }+2 }{ 2 } =1\)
\(\Rightarrow { x }_{ 1 }=3\quad and\quad { y }_{ 1 }=0\)
\(So\quad co-ordinates\quad of\quad focus\quad are\quad (3,0)\)
\(Now\quad PS=PM\Rightarrow { PS }^{ 2 }={ PM }^{ 2 }\)
\(\Rightarrow { \left( x-3 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left[ \frac { x-y+1 }{ \sqrt { { \left( 1 \right) }^{ 2 }+{ -1 }^{ 2 } } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+9-6x+{ y }^{ 2 }\)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }+1-2xy+2x-2y }{ 2 }\)
\(\Rightarrow 2({ x }^{ 2 }+{ y }^{ 2 }-6x+9)={ x }^{ 2 }{ y }^{ 2 }+1-2xy+2x-2y\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-14x+2y+2xy+17=0.\)

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the circle passes through (0, 0),
(0 – h)² + (0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)² + (0 – k)² = h² + k² … (1)
(0 – h)² + (b – k)² = h² + k² … (2)
From equation (1), we obtain
a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
\(\text { However, } a \neq 0 ; \text { hence, }(a-2 h)=0 \Rightarrow h=\frac{a}{2} \text { . }\)
\(\text { From equation (2), we abtain }\)
\(h^{2}+b^{2}-2 b k+k^{2}=h^{2}+k^{2}\)
\(\Rightarrow b^{2}-2 b k=0\)
\(\Rightarrow b(b-2 k)=0\)
\(\Rightarrow b=0 \text { or }(b-2 k)=0\)
\(\text { However, } b \neq 0 \text { ; hence, }(b-2 k)=0 \Rightarrow k=\frac{b}{2} \text { . }\)


Thus, the equation of the required circle is
\(\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2} \)
\(\Rightarrow\left(\frac{2 x-a}{2}\right)^{2}+\left(\frac{2 y-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4} \)
\(\Rightarrow 4 x^{2}-4 a x+a^{2}+4 y^{2}-4 b y+b^{2}=a^{2}+b^{2} \)
\(\Rightarrow 4 x^{2}+4 y^{2}-4 a x-4 b y=0 \)
⇒ x² + y²- ax - by = 0
which is required equation of circle.

Here foci are (0, ±\(\sqrt10\)) which lie on y-axis.
So the equation of hyperbola in standard form is \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ foci (0, ± c) is (0, ± \(\sqrt10\))  a = \(\sqrt10\)
We know that c² = a² + b²
∴ (\(\sqrt10\))² = a² + b² ⇒ b² = 10 - a²
Since the hyperbola passes through (2, 3)
∴ \(\frac { 9 }{ a^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } \)=1
⇒ \(\frac { 9 }{ a^{ 2 } } -\frac { 4 }{ 10-{ a }^{ 2 } } \)=1
⇒ \(\frac { 9(10-{ a }^{ 2 })-4{ a }^{ 2 }={ a }^{ 2 }(10-{ a }^{ 2 }) }{ { a }^{ 2 }(10-{ a }^{ 2 }) } \)
⇒ a⁴ - 23a² + 90 = 0
⇒ a⁴ - 18a² - 5a² + 90 = 0
⇒ (a² - 18) (a² - 5) = 0
⇒ a² = 18 or a² = 5
When a²= 18 then b²= 10-18=-8 (which is not possible)
When a² = 5 then b² = 10 - 5 = 5
Thus required equation of hyperbola is
\(\frac { { y }^{ 2 } }{ 5 } -\frac { { x }^{ 2 } }{ 5 } \)=1