Table of Contents
How do you find the equation of a circle touching the y axis?
We will learn how to find the equation of a circle touches y-axis. The equation of a circle with centre at (h, k) and radius equal to a, is (x – h)2 + (y – k)2 = a2. When the circle touches y-axis i.e., h = a.
Which touches both the axes at a distance of 6 units from the origin?
(i) which touches both the axes at a distance of 6 units from the origin. A circle touches the axes at the points (±6, 0) and (0, ±6). So, a circle has a centre (±6, ±6) and passes through the point (0, 6). We know that the radius of the circle is the distance between the centre and any point on the radius.
Which of the following is the equation of a circle centered at the origin whose diameter is 6 units?
So, if the center is (0,0) and the radius is 6, an equation of the circle is: (x-0)2 + (y-0)2 = 62.
What is the center of the circle ⭕️ in the third quadrant?
Since the circle ⭕️ touches both the axes in the 3rd quadrant, the center of the circle has to be (-p, -p) and it’s radius p (>0). The unknown value p has to be determined.The distance of the center from the given line = (-3p+ 4p+ 8)/v (9+25) = (8+p)/5 = p, since the line touches the given line also. ==> p = 2.
How do you find the tangent of a circle?
Find the equation of a circle which touches both the axes and the line 3x−4y+8=0 and lies in the third quadrant. Let ‘a’ be the radius of the circle. Given that the line 3x−4y+8=0 touches the circle. ∴3x−4y+8=0 is tangent to the circle.
How do you find the center of a circle in R?
Since the circle touch both the axis and is in the third quadrant, it must have center ( − r, − r), where r is the center’s distance from X-axis and Y-axis too, and also r is the radius of the circle. So there is only one unknown r, and we need to find this r.
Where does the origin of a circle lie on the graph?
Because the circle is touching both axes from inside the third quadrant, and since a circle is a set of points equidistant from one central point, we know that the origin of the circle in question lies on the line y = x.