[[1]] is the equation to the circle which touches the axis of X at distance 3 from the origin and intercepts at a distance 6 on the axis of Y.

Question

Infinity Learn · Accepted Answer

We can draw a circle on the XY plane with center C such that, the circle touches the X-axis at (3,0)and intercepts Y-axis at M and N. We can also draw the perpendicular from the Y-axis to the center and it meets the Y-axis at point P.
We are given that the distance the circle intercept on the Y axis is 6. So, we can write,
MN=6
As CM=CN=r, NCM is an isosceles triangle and as PC is perpendicular to NM, P is the midpoint of NM. So, we can write,

PM = NP = \frac{MN}{2} = \frac{6}{2} = 3

As the point where the circle touches the X axis is 3 units away from the Y axis, the center is also 3 units away from the Y axis. So, we get,

PC = 3

Now consider the right triangle PMC. By Pythagoras theorem, we can write,

M C^{2} = P M^{2} + P C^{2}

We have PC = 3,PM = 3 and MC=r as MC is a radius of the circle. Using this in the above equation, we get,

r^{2} = 3^{2} + 3^{2} = 9 + 9 = 18

r = \sqrt{18} = 3 \sqrt{2}

From the figure, the coordinate of the center is the same as the radius. So, the center of the circle is (3,

3 \sqrt{2})

The equation of the circle is given by, (x−x0)2+(y−y0)2=r2
Substituting, with the values of center and radius, we get,
(x−3)2+(y−

3 \sqrt{2})

2=(

3 \sqrt{2}

)2
x2−6x+9+y2−6

\sqrt{3}

y+18=18

x^{2} + y^{2} - 6 x - 6 \sqrt{3} y + 9 = 0

Therefore, the required equation of the circle is

x^{2} + y^{2} - 6 x - 6 \sqrt{3} y + 9 = 0

.

[[1]] is the equation to the circle which touches the axis of X at distance 3 from the origin and intercepts at a distance 6 on the axis of Y.

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