Đáp án:
26) A. 27) D
Giải thích các bước giải:
26)
$\begin{array}{l}
A.{x^2} + {y^2} - 10x + 2y + 1 = 0\\
\Leftrightarrow {(x - 5)^2} + {(y + 1)^2} = 25\\
\Rightarrow \left\{ \begin{array}{l}
I(5; - 1)\\
R = 5
\end{array} \right. \Rightarrow d\left( {I,Oy} \right) = 5 = R \Rightarrow \\
B.{x^2} + {y^2} + x + y - 3 = 0\\
\Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {y + \dfrac{1}{2}} \right)^2} = \dfrac{7}{2}\\
\Rightarrow \left\{ \begin{array}{l}
I\left( {\dfrac{{ - 1}}{2};\dfrac{{ - 1}}{2}} \right)\\
R = \dfrac{{\sqrt {14} }}{2}
\end{array} \right. \Rightarrow d\left( {I,Oy} \right) = \dfrac{1}{2} \ne \dfrac{{\sqrt {14} }}{2} \Rightarrow S\\
C.{x^2} + {y^2} - 1 = 0\\
\Leftrightarrow {x^2} + {y^2} = 1\\
\Leftrightarrow \left\{ \begin{array}{l}
I\left( {0;0} \right)\\
R = 1
\end{array} \right. \Rightarrow d\left( {I,Oy} \right) = 0 \ne 1 \Rightarrow S\\
D.{x^2} + {y^2} - 4y - 5 = 0\\
\Leftrightarrow {x^2} + {\left( {y - 2} \right)^2} = 9\\
\Rightarrow \left\{ \begin{array}{l}
I\left( {0;2} \right)\\
R = 3
\end{array} \right. \Rightarrow d\left( {I,Oy} \right) = 0 \ne 3 \Rightarrow S
\end{array}$
Vậy đáp án đúng là A.
27)
$\begin{array}{l}
\left( C \right):{x^2} + {y^2} - 9 = 0 \Rightarrow \left\{ \begin{array}{l}
I\left( {0;0} \right)\\
R = 3
\end{array} \right.\\
d\left( {I,\Delta } \right) = \dfrac{{\left| {4.0 + 3.0 + m} \right|}}{{\sqrt {{4^2} + {3^2}} }} = \dfrac{{\left| m \right|}}{5}\\
\Leftrightarrow R = 3 = \dfrac{{\left| m \right|}}{5} \Leftrightarrow m = \pm 15 \Rightarrow D
\end{array}$
