Find the equations of the circle passing through (-4, 3) and touching the lines x + y = 2 and x - y

Q: Why are there two angle bisectors? Does the circle have to be on one specific one?

Great question! When two lines intersect, they create four angles. The angle bisectors are the lines that cut these angles in half. Any point on these bisectors is equidistant from the two lines. However, a circle can only be tangent to the lines in one specific region at a time. We use the given point $P$ to 'lock' the circle into the correct region, which dictates which bisector we must use.

Q: Why did we choose $y=0$ as the bisector instead of $x=2$?

Look at the coordinates of our point $P(-4, 3)$. The line $x=2$ is a vertical wall. Our point $P$ is at $x=-4$, which is to the left of this wall. If the center were on $x=2$, the circle would be trapped in a region that doesn't contain $P$. By visualizing the position of $P$ relative to the intersection $(2, 0)$, we can see that $P$ lies in the region bisected by the x-axis ($y=0$).

Q: Is there a way to solve this without assuming the center is $(a, 0)$?

You could use the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, but that introduces three variables ($g, f, c$) and requires solving a system of equations involving tangency conditions and the point $P$. It is mathematically valid but algebraically exhausting! Using the geometric property of the center lying on the bisector reduces the problem to a single variable, which is the hallmark of an elegant JEE solution.

Question

Find the equations of the circle passing through $(-4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced path. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a circle dancing between two lines.

Coordinate geometry is often perceived as a dry landscape of formulas, but I want you to see it as a canvas. We have two lines, $L_1: x + y - 2 = 0$ and $L_2: x - y - 2 = 0$, intersecting at a point $(2, 0)$.

We have a point $P(-4, 3)$ floating in the plane. Our mission is to construct a circle that passes through $P$ and kisses both lines tangentially.

### The Locus of the Center

Before we touch a single algebraic symbol, let us visualize. If a circle is tangent to two lines, its center must be equidistant from both, which is the fundamental definition of a circle's radius.

If the distance to $L_1$ is $r$ and the distance to $L_2$ is $r$, then the center $(h, k)$ must satisfy the condition that the perpendicular distance to both lines is equal. This condition defines the **angle bisectors** of the two lines.

Mathematically, we find these by setting the distance formulas equal:

$$\frac{|x + y - 2|}{\sqrt{1^2 + 1^2}} = \frac{|x - y - 2|}{\sqrt{1^2 + (-1)^2}}$$

When we strip away the absolute values, we get two beautiful, perpendicular lines: $y = 0$ (the x-axis) and $x = 2$. These are the only two paths where our circle's center can possibly reside.

### The Strategic Choice

Now, we must choose. We have two candidates for our center's locus: the vertical line $x = 2$ or the horizontal line $y = 0$.

Look at our point $P(-4, 3)$. It sits at $x = -4$. The line $x = 2$ acts as a boundary; if our center were on $x = 2$, the circle would be confined to the region to the right of that line.

Since $P$ is clearly to the left, the center must lie on the x-axis, $y = 0$. We have successfully reduced our center from a coordinate pair $(h, k)$ to a single variable point $C(a, 0)$.

### The Algebra of Tangency

With the center at $C(a, 0)$, the radius $r$ is simply the distance from $C$ to either line. Using $x + y - 2 = 0$, we find:

$$r = \frac{|a + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{|a - 2|}{\sqrt{2}}$$

This gives us the equation of our circle: $(x - a)^2 + y^2 = r^2$. Substituting our expression for $r^2$, we get:

$$(x - a)^2 + y^2 = \frac{(a - 2)^2}{2}$$

This equation represents the family of all circles centered on the x-axis that are tangent to our two lines. We now need the specific circles that pass through $P(-4, 3)$.

### The Moment of Truth

We substitute $x = -4$ and $y = 3$ into our circle equation. This is the moment where the physics of the problem meets the rigor of the math:

$$(-4 - a)^2 + 3^2 = \frac{(a - 2)^2}{2}$$

Expanding this, we get $(a + 4)^2 + 9 = \frac{(a - 2)^2}{2}$. Multiplying by 2 to clear the fraction, we arrive at:

$$2(a^2 + 8a + 16 + 9) = a^2 - 4a + 4$$

$$2a^2 + 16a + 50 = a^2 - 4a + 4$$

Rearranging everything to one side, we find our quadratic equation:

$$a^2 + 20a + 46 = 0$$

Using the quadratic formula, we solve for $a$:

$$a = \frac{-20 \pm \sqrt{400 - 184}}{2} = -10 \pm \sqrt{54}$$

### Conclusion: The Two Solutions

We have found two values for $a$. This makes perfect sense, as there is a smaller circle nestled closer to the intersection point $(2, 0)$ and a larger circle that wraps around, both satisfying the tangency and the point constraint.

By substituting these values of $a$ back into the general equation, we obtain the final equations of our circles.

Remember, in JEE Advanced, the math is just the language; the geometry is the story. You have successfully navigated the locus, chosen the path, and solved the quadratic. You are not just calculating; you are constructing.

Animated Solution for Mathematics - Circles: Find the equations of the circle passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

Visualized Solution

Visualizing the Geometry

The Angle Bisector Principle

Selecting the Correct Bisector

Expressing the Radius $r$

Applying the Point Constraint

Expanding the Equation

Forming the Quadratic Equation

Solving for $a$

Writing the Equations of the Circles

Summary and Key Takeaways

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Locus of the Center

The Strategic Choice

The Algebra of Tangency

The Moment of Truth

Conclusion

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equations of the circle passing through (−4,3) and touching the lines x+y=2 and x−y=2.