A circle touches the line y = x at a point P such that OP = 4√2, where O is the origin. The circle c

Q: Wait, why did we reject the center $(1, -9)$? The math gave us two possibilities!

Great catch! In coordinate geometry, algebraic solutions often yield multiple candidates. We use the 'interior point' constraint to filter out the extraneous solution. The point $(-10, 2)$ must satisfy the inequality $(x - h)^2 + (y - k)^2 < r^2$. Testing both centers is the only way to be certain which one actually contains the point in its interior.

Q: Why is the normal line $x + y = -8$?

The tangent is $y = x$, which has a slope of $1$. The normal must be perpendicular to it, so its slope is $-1$. Since it passes through $P(-4, -4)$, the equation is $y - (-4) = -1(x - (-4))$, which simplifies to $y + 4 = -x - 4$, or $x + y = -8$.

Q: Can I use the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$ instead?

Absolutely! You can substitute the center $(-9, 1)$ and radius $r = \sqrt{50}$ into the standard form $(x - h)^2 + (y - k)^2 = r^2$ and expand it. It will lead you to the same result. The general form is just a different way of expressing the same geometric truth.

Question

A circle touches the line $y = x$ at a point $P$ such that $OP = 4\sqrt{2}$, where $O$ is the origin. The circle contains the point $(-10, 2)$ in its interior and the length of its chord on the line $x + y = 0$ is $6\sqrt{2}$. Determine the equation of the circle.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving a problem; we are reconstructing a circle from the whispers of its geometric properties.

We are given a circle that kisses the line $y = x$ at a point $P$, and we know the distance from the origin to this point is $4\sqrt{2}$. Since $P$ lies on $y = x$, its coordinates must be of the form $(x_1, x_1)$.

The distance formula tells us that:

$$\sqrt{x_1^2 + x_1^2} = |x_1|\sqrt{2}$$

Equating this to $4\sqrt{2}$, we find $|x_1| = 4$. Thus, $P$ could be $(4, 4)$ or $(-4, -4)$. We shall proceed with $P(-4, -4)$ as our primary path.

### The Normal and the Center

The radius is always perpendicular to the tangent at the point of contact. This is the 'Normal Line'. Since our tangent is $y = x$ (with slope $m = 1$), the normal line must have a slope of $-1$.

For $P(-4, -4)$, the equation of our normal line becomes:

$$y - (-4) = -1(x - (-4)) \Rightarrow x + y = -8$$

The center of our circle, $C(h, k)$, must reside on this line. This gives us our first crucial constraint:

$$h + k = -8$$

### The Chord and the Pythagoras Connection

Next, we encounter the chord on the line $x + y = 0$. If we drop a perpendicular from the center $C(h, k)$ to this chord, it bisects the chord.

The distance $d$ from the center to the line $x + y = 0$ is given by:

$$d = \frac{|h + k|}{\sqrt{1^2 + 1^2}}$$

Substituting our constraint $h + k = -8$, we get:

$$d = \frac{|-8|}{\sqrt{2}} = 4\sqrt{2}$$

The half-chord length is $\frac{6\sqrt{2}}{2} = 3\sqrt{2}$. By the Pythagorean theorem, $r^2 = d^2 + (	ext{half-chord})^2$:

$$r^2 = (4\sqrt{2})^2 + (3\sqrt{2})^2 = 32 + 18 = 50$$

### The Final Filter

We know the center $C(h, k)$ is at a distance $r$ from $P(-4, -4)$. Thus:

$$(h + 4)^2 + (k + 4)^2 = 50$$

Substituting $k = -8 - h$ into the equation:

$$(h + 4)^2 + (-8 - h + 4)^2 = 50 \Rightarrow 2(h + 4)^2 = 50$$

This yields $(h + 4)^2 = 25$, so $h + 4 = \pm 5$. This gives two possible centers: $C_1(-9, 1)$ and $C_2(1, -9)$.

We check the point $Q(-10, 2)$ which lies in the interior. For $C_1(-9, 1)$, the squared distance $CQ^2$ is:

$$(-10 - (-9))^2 + (2 - 1)^2 = (-1)^2 + 1^2 = 2$$

Since $2 < 50$, this point is inside the circle. Thus, the center is $C(-9, 1)$.

### The Equation

With the center at $(-9, 1)$ and $r^2 = 50$, the equation of our circle is:

$$(x + 9)^2 + (y - 1)^2 = 50$$

Expanding this, we get $x^2 + 18x + 81 + y^2 - 2y + 1 = 50$. Simplifying, we arrive at the final form:

**$$x^2 + y^2 + 18x - 2y + 32 = 0$$**

Animated Solution for Mathematics - Circles: A circle touches the line $y = x$ at a point $P$ such that $O P = 42$ , where $O$ is the origin. The circle contains the point $(- 10, 2)$ in its interior and the length of its chord on the line $x + y = 0$ is $62$ . Determine the equation of the circle.

Visualized Solution

Visualizing the Setup

Finding Point $P$

The Normal Line Equation

Distance to Chord Line

Calculating the Radius

Solving for the Center

Interior Point Verification

Final Equation of the Circle

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal and the Center

The Chord and the Pythagoras Connection

The Final Filter

The Equation

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