Trắc nghiệm Toán 10 Bài 3. Đường tròn trong mặt phẳng tọa độ có đáp án !!

A. I(–1; 3), R = 4;           

B. I(1; –3), R = 4;           

C. I(1; –3), R = 16;         

D. I(–1; 3), R = 16.

A. (x + 1)² + (y – 5)² = 26;                  

B. (x + 1)² + (y – 5)² = $\sqrt{26}$ ;              

C. (x – 1)² + (y + 5)² = 26;                  

D. (x – 1)² + (y + 5)² = $\sqrt{26}$

A. (C): (x + 6)² + (y – 7)² = 9;             

B. (C): (x + 6)² + (y – 7)² = 81;           

C. (C): (x + 6)² + (y – 7)² = 89;           

D. (C): (x + 6)² + (y – 7)² = $\sqrt{89}$

A. (x + 2)² + (y – 3)² = 4;          

B. (x + 2)² + (y – 3)² = 9;          

C. (x – 2)² + (y + 3)² = 4;          

D. (x – 2)² + (y + 3)² = 9.

A. I(3; –1), R = 4;           

D. I(–3; 1), R = 2.

A. I(0; 0);              

B. I(1; 0);              

C. I(3; 2);              

D. I(1; 1).

A. 4x² + y² – 10x – 6y – 2 = 0;            

B. x² + y² – 2x – 8y + 20 = 0;              

C. x² + 2y² – 4x – 8y + 1 = 0;              

D. x² + y² – 4x + 6y – 12 = 0.

A. I(–8; 4), R = $\sqrt{91}$ ;                

B. I(8; –4), R = $\sqrt{91}$ ;                

C. I(–8; 4), R = $\sqrt{69}$ ;                

D. $I(-\frac{1}{2};\frac{1}{4}),R=1$

A. (x + 3)² + (y – 1)² = $\sqrt{5}$ ;                

B. (x – 3)² + (y + 1)² = $\sqrt{5}$ ;                

C. (x – 3)² + (y + 1)² = 5;          

D. (x + 3)² + (y – 1)² = 5.

A. m ∈ ℝ;              

A. 5;            

B. 7;            

A. (x – 2)² + (y + 2)² = 25;                  

B. (x + 5)² + (y + 1)² = 16;                  

C. (x + 2)² + (y + 2)² = 9;          

D. (x – 1)² + (y + 3)² = 25.

A. 4x – 3y + 5 = 0; 4x – 3y – 45 = 0;            

B. 4x + 3y + 5 = 0; 4x + 3y + 3 = 0;              

C. 4x + 3y + 29 = 0;                  

D. 4x + 3y + 29 = 0; 4x + 3y – 21 = 0.

A. m = 2;              

B. m = –1;             

C. m = 1;               

D. m = –2.

A. (3; 0);               

B. (–3; 0);             

C. (0; 3);               

D. (0; –3).