Q.Equation of a circle which passes through $(3, 6)$ and touches the axes is
(A) $x^2 + y^2 + 6x + 6y + 3 = 0$
(B) $x^2 + y^2 - 6x - 6y - 9 = 0$
(C) $x^2 + y^2 - 6x - 6y + 9 = 0$
(D) none of these

The key idea is that a circle touching both coordinate axes has its center at $(\pm r, \pm r)$ and a radius of $r$.

Since the point $(3, 6)$ lies in the first quadrant, the circle must also be in the first quadrant. Therefore, its center is $(r, r)$ and its radius is $r$.

1. The standard equation of such a circle is $(x - r)^2 + (y - r)^2 = r^2$.
2. Substitute the given point $(3, 6)$ into the equation: $(3 - r)^2 + (6 - r)^2 = r^2$ $9 - 6r + r^2 + 36 - 12r + r^2 = r^2$ $r^2 - 18r + 45 = 0$
3. Solve the quadratic equation for $r$: $r = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(45)}}{2(1)}$ $r = \frac{18 \pm \sqrt{324 - 180}}{2}$ $r = \frac{18 \pm \sqrt{144}}{2}$ $r = \frac{18 \pm 12}{2}$ This gives two possible values for $r$: $r_1 = \frac{18 + 12}{2} = 15$ and $r_2 = \frac{18 - 12}{2} = 3$. …

When a circle touches both the x-axis and the y-axis, a very specific relationship exists between its center and its radius. Imagine such a circle: the perpendicular distance from its center to the x-axis is its radius, and similarly for the y-axis. This means the absolute value of both the x-coordinate and the y-coordinate of the center must be equal to the radius.

Since the given point $(3, 6)$ has positive coordinates, it lies in the first quadrant. For a circle to pass through this point and touch both axes, it must also be situated in the first quadrant. Consequently, its center $(h, k)$ will have positive coordinates, and its radius $r$ will be positive. This implies $h = r$ and $k = r$.

Given that the circle touches both axes in the first quadrant, its center is $(r, r)$. Substituting this into the standard equation, we get the specific form for such circles:

$(x - r)^2 + (y - r)^2 = r^2$

Now, let's use the information that the circle passes through the point $(3, 6)$ to find the possible values of $r$.

1. Substitute the given point into the circle's equation.

   The circle passes through $(3, 6)$, so these coordinates must satisfy the equation:

   $(3 - r)^2 + (6 - r)^2 = r^2$

2. Expand and simplify the equation to solve for $r$.

   Expand the squared terms:

   $(9 - 6r + r^2) + (36 - 12r + r^2) = r^2$

   Combine like terms:

   $2r^2 - 18r + 45 = r^2$

   Rearrange into a standard quadratic form:

   $r^2 - 18r + 45 = 0$

3. Solve the quadratic equation for $r$.

   We can solve this quadratic equation by factoring or using the quadratic formula. Let's try factoring:

   We need two numbers that multiply to $45$ and add up to $-18$. These numbers are $-3$ and $-15$.

   $(r - 3)(r - 15) = 0$

   This gives two possible values for the radius $r$:

   $r = 3$ or $r = 15$

   This means there are two distinct circles that satisfy the given conditions.

4. Formulate the equation for each possible value of $r$.

   • Case 1: $r = 3$ The center of the circle is $(3, 3)$ and the radius is $3$. …