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
14. A circle has centr e^candequationx^2+y^2-6x+14y+49=0(a)Find the centre of the circle.(...)(2)(b)Find the radius of the circle(2) 相关知识点: 试题来源: 解析 (a)(x-3y)^2-9^2+14y+4y=0.99-0.92-0.96+3.29)=99(3,7)(b)3 反馈 收藏...
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...
the equation of the circle is(x -4)2+-|||-(y-4)2=16or(x-1)2+(y+1)2=1.-|||-Method 2:-|||-Let the equation of the circle be-|||-(x-a)2+(y-b)2=r2-|||-Since the circle is tangent to both axis-|||-a2=b2=r2-|||-(1)-|||-Since the center of the circle ...
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^...
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...
Steps for Finding the Tangent of a Circle Step 1: Determine which length is missing in the figure. Step 2: Solve the equation or use the Pythagorean Theorem to find the missing length. Equations and Definitions for Finding the Tangent of a Circle Tangent Line: a straight line that touche...
Center and Radius of a CircleA circle is a flat geometry with a uniform rounded periphery. The periphery is known as the circumference of the circle. Each point on the circumference of a circle is equidistance from a fixed point called the center. And the distance ...
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 ...
The circle is a geometric shape that has all its points at a constant distance from a point. Specifying this point (center of the circle) and the constant distance to the circle's points (the radius) is possible to construct the mathematical equation for the circle, ...