Find an equation of the circle that has center (-3,3)and passes through (4,-1). 相关知识点: 试题来源: 解析 x^2+y^2+6x-6y=47 r=√((x_2-x_1)^2+(y_2-y_1)^2) (x+3)^2+(y-3)^2=(√5)^2 (x^2+y^2+6x-6y-6x-11)/(x^2+y^2+6x.6y=0.7) M ...
Find the equation of the circle passing through the points (4 3) (-2 -5) and (5 2).我真的对这题完全没辙了.马上考试了. 相关知识点: 试题来源: 解析 想要做圆方程,首先要找圆心A(4,3) B(-2,-5) C(5,2)过AC的直线为 y=-x 7 ,AC的中点为(4.5,2.5),则AC的中垂线为 y=x-2过BC...
百度试题 结果1 题目【题目】Find the equation of the circle that circumscribes the triangle in Erercise.P_1=(-2,1) P_2=(6,-5) P_3=(2,-7) 相关知识点: 试题来源: 解析 【解析】(x-2)^2+(y+2)^2=5^2 反馈 收藏
【解析】Solution:(x-4)^2+(y-4)^2=160r(x-1)^2+ (y+1)^2=1 Method 1:Since the circle is tangent to both axes, the center ofthe circle is on the line.Solving5x-3y=8;x±y=0. x=4;y=4. =4arx=1;y=-1.Therefore the equation of the circle is(x -4)2+(y-4)^2=160r...
1. Find the equation of the circle with(i) Centre at origin and radius 4.(ii) Centre at (-3, -2) and radius 6.(iii) Centre at (2, -3) and radius 5.(iv) Centre at (-3, -3) passing through point(-3,-6) 相关知识点: 试题来源: 解析 (i)x^2+y^2=16(ii)x^2+y^...
the circle passes the origin,we let the equa--|||-tion be x+y+Dx Ey=0.-|||-We know that the circle passes points (a,0),(0,-|||-b).-|||-So a2 Da=0 and 62+Eb=0.-|||-Since a≠0,b≠0,SoD=-a,E=-b.-|||-The equation of the circle is x+y-ax -by =0.优质...
Problem 22:Find the equation of the circle passingthrough the points A(1,2)and B(3,4)and intersectingthe x-axis at two points to form a chord of length of 6.Problem 22:Find the equation of the circle passing through the points A(1, 2) and B(3,4)and intersecting the x-axis at...
百度试题 结果1 题目Find the equation of the circle that circumscribes the triangle in Exercise.P_1=(10,-21)P_2=(-6,-9)P_3=(3,3) 相关知识点: 试题来源: 解析 (x-(13)2)^2+ (y+9)^2=((25)2)^2 反馈 收藏
Let the equation of the general form of the required circle be x² + y² +2gx+ 2fy + c= 0...(i)According to the problem, the above equation of the circle passes throughthe points (1, 0), (-1, 0) and (0, 1). Therefore,1 + 2g+ c= 0...(ii)1 -2g+ C=0...(iii...
Let the equation of the required circle be .Since the circle passes through points and .Since the centre of the circle lies on line ,From equations (1) and (2), we obtainOn solving equations (3) and (4), we obtain and .On substituting the values of and in equation (1), we obtain...