25x² + 9y² = 225

\(\left( 0,\frac { 3 }{ 2 } \right) \); x=0; 2y+3=0; 6

(4, 0); y = 0; x + 4 = 0; 16

Focus = (0, –3); directrix y = 3
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4ay or x² = – 4ay.
It is also seen that the directrix, y = 3 is above the x-axis, while the focus (0, –3) is below the x-axis. Hence, the parabola is of the form x² = –4ay.
Here, a = 3
Thus, the equation of the parabola is x² = –12y.

Focus (6, 0); directrix, x = –6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y² = 4ax or y² = – 4ax.
It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis. Hence, the parabola is of the form y² = 4ax.
Here, a = 6
Thus, the equation of the parabola is y² = 24x.

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

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

Here h =-a, k =-b and r = \(\\ \sqrt { { a }^{ 2 }-{ b }^{ 2 } } \)
The equation of circle is
(x - h)² + (y - k)² = r²
∴ (x + a)² + (y + b)² = (\(\\ \sqrt { { a }^{ 2 }-{ b }^{ 2 } } \))²
⇒ x² + a² + 2ax + y² + b² + 2by = a² - b²
⇒ x² + y² + 2ax + 2by + 2b² = 0
which is required equation of circle.

Here h = 1, k = 1 and r = \(\sqrt { 2 } \)
The equation of circle is
(x - h)² + (y- k)² = r²
(x - 1)² + (y - 1)² =(\(\sqrt { 2 } \))²
⇒ x²+1-2x+y²+1-2y=2
⇒ x²+y²-2x-2y=0
which is required equation of circle.

Centre of circle is mid-point of end point of diameter.
Here, centre = (2,3)
Let the other end be (x,y)
Then, \(\\ \frac { x+3 }{ 2 } =2,\frac { y+4 }{ 2 } =3\)
= (1, 2)

Centre of given circle x + y + 4x + 5y -\(\\ \frac { 39 }{ 2 } \) = 0 is \(\\ (-2,-\frac { 5 }{ 2 } )\) and area of circle, \(\\ \pi \)r² = 16 \(\\ \pi \)
r²=16
Then, the equation of required circle is \((x+2)^{ 2 }+(y+\frac { 5 }{ 2 } )^{ 2 }=16\)
4x² + 4y² + 16x + 20y - 23 = 0

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

\(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 3 } =1\)

\(\frac { { x }^{ 2 } }{ 121 } +\frac { { y }^{ 2 } }{ 112 } =1\)

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