Phương trình đường tròn

\(\left( C \right):{x^2} + {y^2} = 9\)\( \Rightarrow I\left( {0;0} \right),R = \sqrt 9  = 3\)

Điều kiện để $\left( C \right):{x^2} + {y^2} - 2ax - 2by + c = 0$ là phương trình đường tròn là \({a^2} + {b^2} - c > 0\).

\(\left( C \right):{x^2} + {y^2} + 12x - 14y + 4 = 0\) \( \to \left\{ \begin{array}{l}I\left( { - 6;7} \right)\\R = \sqrt {36 + 49 - 4}  = 9\end{array} \right.\)

\( \to \left( C \right):{\left( {x + 6} \right)^2} + {\left( {y - 7} \right)^2} = 81.\)

Ta có: ${x^2} + {y^2} - 2mx - 4\left( {m - 2} \right)y + 6 - m = 0$ $ \to \left\{ \begin{array}{l}a = m\\b = 2\left( {m - 2} \right)\\c = 6 - m\end{array} \right.$ $ \to {a^2} + {b^2} - c > 0$

$ \Leftrightarrow 5{m^2} - 15m + 10 > 0 \Leftrightarrow \left[ \begin{array}{l}m < 1\\m > 2\end{array} \right..$

Ta có: \({x^2} + {y^2} - 2x + 2my\; + {\rm{ }}10 = 0\) \( \to \left\{ \begin{array}{l}a = 1\\b =  - m\\c = 10\end{array} \right.\) \( \to {a^2} + {b^2} - c > 0\) \( \Leftrightarrow {m^2} - 9 > 0\) \( \Leftrightarrow \left[ \begin{array}{l}m <  - 3\\m > 3\end{array} \right. \) \(\Leftrightarrow m = 4;5 \ldots ;10.\)

$\left( C \right):{\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} = 25$ \( \Leftrightarrow {x^2} - 2x + 1 + {y^2} + 4y + 4 = 25\)$ \Leftrightarrow {x^2} + {y^2} - 2x + 4y - 20 = 0$

\({x^2} + {y^2} + 2mx - 4\left( {m + 1} \right)y + 4{m^2} + 5m + 2 = 0\,\left( 1 \right)\)

Có \(a =  - m,\,\,b = 2\left( {m + 1} \right),\) \(c = 4{m^2} + 5m + 2\)

(1) là phương trình đường tròn \( \Leftrightarrow {a^2} + {b^2} - c > 0\)

\(\begin{array}{l} \Leftrightarrow {\left( { - m} \right)^2} + 4{\left( {m + 1} \right)^2} - \left( {4{m^2} + 5m + 2} \right) > 0\\ \Leftrightarrow {m^2} + 4\left( {{m^2} + 2m + 1} \right) - 4{m^2} - 5m - 2 > 0\\ \Leftrightarrow {m^2} + 4{m^2} + 8m + 4 - 4{m^2} - 5m - 2 > 0\\ \Leftrightarrow {m^2} + 3m + 2 > 0\\ \Leftrightarrow \left( {m + 1} \right)\left( {m + 2} \right) > 0\\ \Leftrightarrow \left[ \begin{array}{l}m >  - 1\\m <  - 2\end{array} \right.\end{array}\)

Ta có $\left( C \right):{x^2} + {y^2} - 6x + 2y + 6 = 0$ $ \to a = \dfrac{{ - 6}}{{ - 2}} = 3,b = \dfrac{2}{{ - 2}} =  - 1,c = 6$

$ \to I\left( {3; - 1} \right),R = \sqrt {{3^2} + {{\left( { - 1} \right)}^2} - 6}  = 2.$

\(\left( C \right):{x^2} + {y^2} - 4x + 6y - 12 = 0\)

\( \to a = 2,b =  - 3,c =  - 12\)

\( \to I\left( {2; - 3} \right),R = \sqrt {4 + 9 + 12}  = 5\)

Ta có:\(\left( C \right):2{x^2} + 2{y^2} - 8x + 4y - 1 = 0\) \( \Leftrightarrow {x^2} + {y^2} - 4x + 2y - \dfrac{1}{2} = 0\) \( \to \left\{ \begin{array}{l}a = 2,b =  - 1\\c =  - \dfrac{1}{2}\end{array} \right.\)

\( \to I\left( {2; - 1} \right),R = \sqrt {4 + 1 + \dfrac{1}{2}}  = \dfrac{{\sqrt {22} }}{2}\)

\(\left( C \right):16{x^2} + 16{y^2} + 16x - 8y - 11 = 0\) \( \Leftrightarrow {x^2} + {y^2} + x - \dfrac{1}{2}y - \dfrac{{11}}{{16}} = 0\)

\( \to \left\{ \begin{array}{l}I\left( { - \dfrac{1}{2};\dfrac{1}{4}} \right)\\R = \sqrt {\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{11}}{{16}}}  = 1.\end{array} \right.\)

$\left( C \right):{x^2} + {y^2} - 10x + 1 = 0 \to I\left( {5;0} \right)$ $ \to d\left( {I;Oy} \right) = \left| 5 \right| = 5$

$\left( C \right):{x^2} + {y^2} + 5x + 7y - 3 = 0 \to I\left( { - \dfrac{5}{2}; - \dfrac{7}{2}} \right)$  $ \to d\left( {I;Ox} \right) = \left| { - \dfrac{7}{2}} \right| = \dfrac{7}{2}$

Đường tròn \(\left( C \right)\) có phương trình \({x^2} + {y^2} - 8x - 6y + 21 = 0\) sẽ có tâm \(I\left( {4;3} \right)\) bán kính \(R = \sqrt {{4^2} + {3^2} - 21}  = 2\).

Ta có \(MI = \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( {2 - 3} \right)}^2}}  = 1 < R = 2 \)

\(\Rightarrow M\) nằm trong \(\left( C \right)\)

$\left( C \right):\left\{ \begin{array}{l}I\left( {1; - 5} \right)\\R = OI = \sqrt {26} \end{array} \right.$ $ \to \left( C \right):{\left( {x - 1} \right)^2} + {\left( {y + 5} \right)^2} = 26$

$\left( C \right):\left\{ \begin{array}{l}I\left( { - 2;3} \right)\\R = IM = \sqrt {{{\left( {2 + 2} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}}  = \sqrt {52} \end{array} \right.$ $ \to \left( C \right):{\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2} = 52$

$\left( C \right):\left\{ \begin{array}{l}I\left( {1;2} \right)\\R = 3\end{array} \right.$ $ \to \left( C \right):{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = 9$ $ \Leftrightarrow {x^2} + {y^2} - 2x - 4y - 4 = 0$

$\left( C \right):\left\{ \begin{array}{l}I\left( {2; - 3} \right)\\R = \dfrac{1}{2}AB = \dfrac{1}{2}\sqrt {{{\left( {1 - 3} \right)}^2} + {{\left( { - 5 + 1} \right)}^2}}  = \sqrt 5 \end{array} \right.$  $ \to \left( C \right):{\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 5$

$\left( C \right):\left\{ \begin{array}{l}I\left( {4;3} \right)\\R = IA = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( {3 - 1} \right)}^2}}  = \sqrt {13} \end{array} \right.$ $ \to \left( C \right):{\left( {x - 4} \right)^2} + {\left( {y - 3} \right)^2} = 13$

$ \Leftrightarrow {x^2} + {y^2} - 8x - 6y + 12 = 0.$

\(I\left( {a;0} \right) \to IA = IB\) \( \Leftrightarrow {\left( {a - 1} \right)^2} + {1^2} = {\left( {a - 5} \right)^2} + {3^2}\) \( \Leftrightarrow {a^2} - 2a + 1 + 1 = {a^2} - 10a + 25 + 9\)

\( \Leftrightarrow 8a - 32 = 0 \Leftrightarrow a = 4\)

\( \to \left\{ \begin{array}{l}a = 4\\I\left( {4;0} \right)\\{R^2} = I{A^2} = 10\end{array} \right.\)

Vậy đường tròn cần tìm là: \({\left( {x - 4} \right)^2} + {y^2} = 10.\)