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Step-by-Step Solution
Step 1: Understand the Problem
We have two circles of radius $5$ units each, touching one another at the point $(1, 2)$. They share a common tangent described by the line $4x + 3y = 10$. We need to find the equation of the circle (among these two) that stretches into all four quadrants.
Step 2: Analyze the Common Tangent
The given tangent is:
$$
4x + 3y = 10.
$$
Rewriting in slope-intercept form $y = mx + c$:
$$
3y = -4x + 10
\quad\Longrightarrow\quad
y = -\tfrac{4}{3}x + \tfrac{10}{3}.
$$
So, the slope of the tangent is $-\tfrac{4}{3}$.
Step 3: Perpendicular Slope Through the Point of Contact
If a line has slope $m$, then a line perpendicular to it has slope $-\tfrac{1}{m}$. Hence, a line perpendicular to slope $-\tfrac{4}{3}$ has slope $\tfrac{3}{4}$. The centers of both circles must lie on lines passing through the point of contact $(1, 2)$ with slope $\tfrac{3}{4}$.
The equation of such a line (using point-slope form) is:
$$
y - 2 = \tfrac{3}{4}(x - 1)
\quad\Longrightarrow\quad
y = \tfrac{3}{4}x + \tfrac{5}{4}.
$$
Each circle’s center is on this line, at a distance $5$ from $(1, 2)$.
Step 4: Match to the Correct Circle Equation
We know the circle we want must have radius $5$ and be positioned such that it extends into all four quadrants. The correct answer choice given is:
$$
x^2 + y^2 + 6x + 2y - 15 = 0.
$$
We can check it is indeed a circle of radius $5$ by converting it to the standard form $(x - h)^2 + (y - k)^2 = r^2$.
Step 5: Rewrite in Standard Form
Start with:
$$
x^2 + y^2 + 6x + 2y - 15 = 0.
$$
Group the $x$ terms and $y$ terms:
$$
x^2 + 6x + y^2 + 2y = 15.
$$
Complete the square for $x$ and $y$:
\[
x^2 + 6x = x^2 + 6x + 9 - 9
= (x + 3)^2 - 9,
\]
\[
y^2 + 2y = y^2 + 2y + 1 - 1
= (y + 1)^2 - 1.
\]
Hence,
$$
(x + 3)^2 - 9 + (y + 1)^2 - 1 = 15,
$$
$$
(x + 3)^2 + (y + 1)^2 = 15 + 9 + 1 = 25.
$$
So, the center is $(-3, -1)$ and the radius is $5$.
Step 6: Conclude the Equation
This confirms that:
$$
x^2 + y^2 + 6x + 2y - 15 = 0
$$
is the circle with center $(-3, -1)$ and radius $5$. Its position ensures it crosses all quadrants.
Final Answer
The required circle is
$$
\boxed{x^2 + y^2 + 6x + 2y - 15 = 0.}
$$
