Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is (x + 3)² + (y –2)² = 16

Here h = k = 0. Therefore, the equation of the circle is x² + y² = r².

\({ x }^{ 3 }+{ y }^{ 2 }-6x-8y+15=0\)

The equation of circle is (x - h)² + (y - k)² = r².. (i)
Since the circle passes through point (4,5) and co-ordinates of centre are (2, 2)
∴ radius of circle
=\(\sqrt { (4-2)^{ 2 }+(5-2)^{ 2 } } =\sqrt { 4+9 } =\sqrt { 13 } \)
Now the equation of required circle is
(x - 2)² + (y - 2)² = (\(\sqrt13\))²
⇒ x²+ 4 - 4x +y² + 4 - 4y = 13
⇒ x² + y² - 4x - 4y - 5 =0.

outside the circle

The given equation of circle is
2x² + 2y² - x = 0
∴ x²+y²-\(\frac { x }{ 2 } \)=0
⇒ \(\left( { x }^{ 2 }-\frac { x }{ 2 } \right) +{ y }^{ 2 }\)=0
⇒ \(\left[ { x }^{ 2 }-\frac { x }{ 2 } +\left( \frac { 1 }{ 4 } \right) ^{ 2 } \right] +{ y }^{ 2 }=0+\left( \frac { 1 }{ 4 } \right) ^{ 2 }\)
⇒ \(\left( x-\frac { 1 }{ 4 } \right) ^{ 2 }+{ y }^{ 2 }=\left( \frac { 1 }{ 4 } \right) ^{ 2 }\)
Comparing it with (x - h)² + (y - k)² = r², we have
h=\(\frac { 1 }{ 4 } \), k=0 and r=\(\frac { 1 }{ 4 } \)
Thus co-ordinates of the centre is \(\left( \frac { 1 }{ 4 } ,0 \right) \) and radius is \(\frac { 1 }{ 4 } \).

The given equation of circle is
x² + y² - 8x + 10y - 12 = 0
∴ (x² - 8x) + (y² + 10y) = 12
⇒ [x² - 8x + (4)²] + [y² + 10y + (5)²]
= 12 + (4)² + (5)²
⇒ (x - 4)² + (y + 5)² = 12 + 16 + 25
⇒ (x - 4)² + (y + 5)² = 53
⇒ (x - 4)² + (y + 5)² =(\(\sqrt { 53 } \))²
Comparing it with (x - h)2 + (y - k)2 = r2, we have
h = 4, k = - 5 and r=\(\sqrt { 53 } \)
Thus co-ordinates of the centre is (4, -5) and radius is \(\sqrt { 53 } \) .

The given equation of circle is
x² + y² - 4x - 8y - 45 = 0
∴ (x² - 4x) + (y² - 8y) = 45
⇒ [x² - 4x + (2)²] + [y² - 8y + (4)²]
= 45 + (2)² + (4)²
⇒ (x - 2)² + (y - 4)² = 45 + 4 + 16
⇒ (x - 2)² + (y - 4)² = 65
⇒ (x - 2)² + (y - 4)² = (\(\sqrt { 65 } \))²
Comparing it with (x- h)² + (y - k)² = r², we have
h = 2, k = 4 and r = \(\sqrt { 65 } \)
Thus co-ordinates of the centre is (2, 4) and radius is \(\sqrt { 65 } \).

The given equation of circle is
(x + 5)² + (y - 3)² = 36
⇒ (x + 5)² + (y - 3)² = (6)²
Comparing it with (x - h)² + (y - k)² = r² we have
h = - 5, k = 3 and r = 6
Thus the coordinates of the centre is (-5, 3) and radius is 6

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.

Here h=\(\frac { 1 }{ 2 } \) , k=\(\frac { 1 }{ 4} \) and r=\(\frac { 1 }{ 12 } \)
The equation of circle is
(x - h)² + (y - k)² = r²
∴ \(\left( x-\frac { 1 }{ 2 } \right) ^{ 2 }+\left( y-\frac { 1 }{ 4 } \right) ^{ 2 }=\left( \frac { 1 }{ 12 } \right) ^{ 2 }\)
⇒ \({ x }^{ 2 }+\frac { 1 }{ 4 } -x+{ y }^{ 2 }+\frac { 1 }{ 16 } -\frac { 1 }{ 2 } y=\frac { 1 }{ 144 } \)
⇒ 144x² + 36 - 144x + 144y² + 9 - 72y = 1
⇒ 144x² + 144y² - 144x - 72y + 44 = 0
⇒ 4(36x² + 36y² - 36x - 18y + 11) = 0
⇒ 36x² + 36y² - 36x -18y + 11 = 0
which is required equation of circle.

Here h = -2, k = 3 and r = 4
The equation of circle is
(x - h)² + (y - k)² = r²
∴ (x + 2)² + (y - 3)² = (4)²
⇒ x² + 4 + 4x + y² + 9 - 6y = 16
⇒ x² + y² + 4x - 6y - 3 = 0
which is required equation of circle.

Here h = 0, k = 2 and r = 2
The equation of circle is
(x - h)² + (y - k)² = r²
∴ (x - 0)² + (y - 2)² = (2)²
x² + y² + 4 - 4y = 4
x² + y² - 4y = 0.
which is required equation of circle.

(i) Given, (x + 5)² +(y -3)² = 36
On comparing the above equation with
(x-h)² +(y - k)² = r², we get
 h = - 5, k = 3 and r = 6
Centre (h,k)= (-5,3) and radius (r)= 6
(ii) Given x² + y² - 4x - 8y - 45 = 0
⇒⇒ (x² - 4x)+(y² - 8y) = 45
⇒⇒ (x² - 4x + 4) + (y² - 8y + 16) = 4 + 16 + 45
⇒(x−2)²+(y−4)²=\(( \sqrt{65} )^2\)
On compare the above equation with
(x - h)² + (y - k)² = r², we get
h = 2, k = 4 and r = \( \sqrt{65} \)
∴Centre (h, k) ≡ (2,4) and radius =\( \sqrt{65} \)
(iii) Centre ≡ (3, - 2),radius = 5
(iv) (- 4, - 5), 7
(v) (1,- 2),\( \sqrt{13} \)
(vi) (- 3, 4),\( \sqrt{33} \)
(vii) (- 4, 5), 5
(viii) (- 5,4),\( \sqrt{77} \)
(ix) Centre ≡ (-1,2/3), radius = 4/3

The equation of the given circle is x² + y² = 25.
x² + y² = 25
⇒ (x – 0)² + (y – 0)² = 5², which is of the form (x – h)² + (y – k)² = r², where h = 0, k = 0, and r = 5.
∴ Centre = (0, 0) and radius = 5
Distance between point (–2.5, 3.5) and centre (0, 0)
\(=\sqrt{(-2.5-0)^{2}+(3.5-0)^{2}} \)
\(=\sqrt{6.25+12.25} \)
\(=\sqrt{18.5} \)
\(=4.3(\text { approx. })<5\)
Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.

Inside the circle.

Let the equation of circle with centre 9h,k) and radius r be
 (x- h)² +(y - k)²= r² ......(i)
Since, circle passes through the points (2,-2) and (3,4), so the points (2,-2) and (3,4) will lie on Eq.(i)
(2 - h)²+ (-2-k)² = r².....(ii)
and  (3 - h)²+(4 - k)² = r²...(iii)
Now, from Eqs (ii) and (iii), we get
(2-h)²+(-2-k)²=(3-h)² +(4-k)²
2h + 12k = 17.....(iv)
Also, given that centre (h,k) lies on x+y =2.So, it will satisfy it
h +k = 2.....(v)
On solving Eqs.(iv) and (v) we get
h= 0.7,k = 13
Now,from Eq,(ii) we get
r² = (2 - 0.7)² + (2 - 13)² = 1.69+10.89 = 12.58
Now, put the values of h,k and r in eq.(i), we get the answer
(x - 0.7)²+(y - 1.3)²=12.58

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.

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 (2, 3)
∴ (2 - h)² + (3 - k)² = r²
⇒ 4 + h² - 4h + 9 + k² - 6k = r²
⇒ h² + k² - 4h - 6k + 13 = r² ...(ii)
Also the circle passes through point (-1, 1)
∴ (-1-h)² + (1- k)² = r²
⇒ 1 + h² + 2h + 1 + k² - 2k = r²
⇒ h² + k² + 2h - 2k + 2 = r²...(iii)
From (ii) and (iii), we have
h² + k² - 4h - 6k + 13 = h² + k² + 2h - 2k + 2
⇒ -6h-4k=-11
⇒ 6h + 4k = 11...(iv)
Since the centre (h, k) of the circle lies on the line x - 3y - 11 = 0
∴ h- 3k-11 = 0
⇒ h - 3k = 11...(v)
Solving (iv) and (v), we have
h=\(\frac { 7 }{ 2 } \) and k=\(\frac { -5 }{ 2 } \)
Putting these values of hand k in (ii), we have
\(\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\frac { -5 }{ 2 } \)=r²
⇒ \(\frac { 49 }{ 4 } +\frac { 25 }{ 4 } \)-14+15+13=r²
⇒ r²=\(\frac { 65 }{ 2 } \)
Thus equation of required circle is
\(\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } \)
⇒ x²+\(\frac { 49 }{ 4 } \)-7x+y²+\(\frac { 25 }{ 4 } \)+5y=\(\frac { 65 }{ 2 } \)
⇒ 4x² +49-28x+4y² +25+20y= 130
⇒ 4x² + 4y² - 28x + 20y - 56 = 0
⇒ 4(x² + y² - 7x + 5y - 14) = 0
⇒ x² + y² - 7x + 5y - 14 =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.