the circle touches x axis at a distance 3 from origin and intercept a distance of 6 on y axis.Find equation of

Question

of the circle.

Answer 1

impossible... you have an ellipse with a major axis of length 2a (6 in the y direction) and a minor axis of length a (3 in the x direction). The equation of such an ellipse in the standard format (x-h)^2 / a^2 + (y-k)^2 / b^2 = 1 would have h and k values of 0 (considering you're starting from the origin), an a value of 3 and a b value of 6, thus resulting in:

x^2 / 3^2 + y^2 / 6^2 = 1 or simply x^2 / 9 + y^2 / 36 = 1

Answer 2

Let try this again. When you said it touches the x-axis at distance 3 from the origin. You mean it touches in only one points. You could mean it touches at (3,0) or (-3,0) but we will assume (3,0) When a curve only touches an axis in only one point that point is on a diameter, so the center includes the point (3,b)
this gives up the following equation and it includes the point (3,0)
(x-3)^2 + (y-b)^2 = r^2
so insert x= 3 and y=0
b^2= r^2
so now we
(x-3)^2 + (y-b)^2 = b^2
Then you said it intersects y at a distance of 6 , we will assume by intercepts you mean it touches in more than one points.
now insert point (0,6)
-3^2 + (6-b)^2 = b^2
9 + 36-12b +b^2= b^2
12b=45
b= 45/12 = 15/4

so we have the following circle

(x-3)^2 + (y-15/4)^2 = 15*15/4*4 = 14 1/16

(x-3)^2 + (y-15/4)^2= 225/14

points (3,0) and (0,6) meet the above equation
(3,0)
0 + (15/4)^2 = 225/16
(0,6)
(0-3)^2 + (24/4-15/4)^2 = 9 + (9/4)^2 = (144+81)/ 16= 225/16

so we have a curve that touches the x axis at (3,0)
it touches the y-axis at (0,6) and (0,1.5)

Answer 3

So are you saying you need the equation of a circle passing through (3,0) and (0,6)? You need more here.

Answer 4

u need concebtrate o what the instructor instructs u.
may be u r busy looking at hot chicks