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...
Here, the centre of the given circle is at(4,-4). As the circle touches both the axes, its radius is 4 units. The equation of the circle is(x-h)²+(y-k)²=r²or,(x-4)²+{y-(-4)}²=4²or,(x-4)²+(y+4)²=16or,x²-8x+16+y²+8y+16=16or,x²...
题目Find the equation of a circle whose center lies on the linex+4=0, and is tangent to the y‑axis at (0,−2). 相关知识点: 试题来源: 解析 (x + 4)= + (y + 2)= = 16(x + 4)= + (y + 2)= = 16 反馈 收藏
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^...
【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...
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...
find the equation of the circle whose centre lies on the line 2x + y -1=0 and which passes throught the points A(-2,0)and B(5,1) 相关知识点: 试题来源: 解析 圆心在直线 2x + y -1=0且过(-2,0)和 B(5,1) 根据题意 圆心还必然在(-2,0)和 B(5,1) 连线的中垂线上 这条...
Solution:x+y-ax-by =0.-|||-Since 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 ...
In Problems, find the standard form of the equation of the circle whose center and radius are given.(h,k)=(-1,-2); r=1 相关知识点: 试题来源: 解析 (x+1)^2+(y+2)^2=1 结果一 题目 B.所给词适当形式填空 51.Liu Hao is the chief engineer of the high-speed railway- (connect)...
In this case, ( r=3√2) and the center point is ( (5,3)). The equation for the circle is ( ((x-(5)))^2+((y-(3)))^2=((3√2))^2). ( ((x-(5)))^2+((y-(3)))^2=((3√2))^2) The circleequation is ( ((x-5))^2+((y-3))^2=18). ( ((x-5))^2+...