Question

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

Answer 1

Hint: We can draw the circle with the given details on the XY plane. We already have the x coordinate of the center. As the circle touches the X-axis, the y coordinate of the center will be equal to the circle’s radius. From the figure, we can find the radius using the intercepts of the Y-axis geometrically. With the center and radius, we can find the equation of the circle.

Complete step by step 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.(3,0)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 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, 

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

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

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

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

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

$r^2=3^2+3^2=9+9=18$

By taking the square root, we get,

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

Thus, we have radius of the circle as $3\sqrt{2}$.

From the figure, y coordinate of the center is same as the radius. So, the center of the circle is $\left(3,3\sqrt{2}\right)$

The equation of the circle is given by, ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=r^2$

Substituting, with the values of center and radius, we get,

${\left(x-3\right)}^2+{\left(y-3\sqrt{2}\right)}^2={\left(3\sqrt{2}\right)}^2$

Opening the brackets, we get,

$x^2-6x+9+y^2-6\sqrt{3}y+18=18$

On simplification, we get,

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

Therefore, the required equation of the circle is $x^2+y^2-6x+-6\sqrt{3}y+9=0$

Note: The concept of two-dimensional geometry is used to solve the problem. We draw the figures using graph theory. Then we used properties of an isosceles triangle to prove P is the mid-point of MN. We used the Pythagoras theorem to find the radius of the circle. Every point on a circle has equal distance from the center and this distance is the radius of the circle. We use this property of the circle to prove the triangle formed is an isosceles triangle.