Let the lengths of intercepts on x-axis and y-axis made by the circle x² + y² + ax + 2ay + c = 0 (a

Q: Wait, why do we use $2\sqrt{g^2 - c}$ for the $x$-intercept? Can't we just set $y=0$ in the circle equation?

You absolutely can! Setting $y=0$ in $x^2 + y^2 + ax + 2ay + c = 0$ gives $x^2 + ax + c = 0$. The roots of this quadratic are the $x$-coordinates of the intercept points. The length of the intercept is the absolute difference of these roots, $|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1x_2} = \sqrt{a^2 - 4c} = 2\sqrt{(a/2)^2 - c}$. Both methods are identical; the formula is just a shortcut!

Q: Why is the slope of the tangent 2? I thought the slope of $x + 2y = 0$ was $1/2$.

Be careful with the rearrangement! The line $x + 2y = 0$ is $y = -\frac{1}{2}x$, so its slope is $-1/2$. For a line to be perpendicular, the product of the slopes must be $-1$. Thus, $m \cdot (-1/2) = -1$, which gives $m = 2$. You were close, but watch those negative signs!

Q: Does it matter which tangent I choose? The formula gives a $\pm$ sign.

Great observation! The $\pm$ sign indicates there are two parallel tangents to the circle with the same slope. Since the question asks for the shortest distance from the origin to *a* tangent, and both tangents are equidistant from the center, you will find that both tangents are actually equidistant from the origin in this specific case. You can pick either one!

Question

Let the lengths of intercepts on $x$-axis and $y$-axis made by the circle $x^2 + y^2 + ax + 2ay + c = 0$ ($a < 0$) be $2\sqrt{2}$ and $2\sqrt{5}$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x + 2y = 0$, is equal to :

Options:

(A) $\sqrt{11}$

(B) $\sqrt{7}$

(C) $\sqrt{6}$

(D) $\sqrt{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the circle equation $x^2 + y^2 + ax + 2ay + c = 0$ with the constraint $a < 0$. The lengths of the intercepts on the $x$ and $y$ axes are $2\sqrt{2}$ and $2\sqrt{5}$, respectively.

Our objective is to find the shortest distance from the origin to a tangent of this circle that is perpendicular to the line $x + 2y = 0$.

### Decoding the Circle

For a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the $x$-intercept length is $2\sqrt{g^2 - c}$ and the $y$-intercept length is $2\sqrt{f^2 - c}$. In our equation, $g = a/2$ and $f = a$.

The $x$-intercept condition gives:
$$2\sqrt{\left(\frac{a}{2}\right)^2 - c} = 2\sqrt{2} \Rightarrow \frac{a^2}{4} - c = 2$$

The $y$-intercept condition gives:
$$2\sqrt{a^2 - c} = 2\sqrt{5} \Rightarrow a^2 - c = 5$$

Subtracting the first equation from the second eliminates $c$:
$$\left(a^2 - c\right) - \left(\frac{a^2}{4} - c\right) = 5 - 2 \Rightarrow \frac{3a^2}{4} = 3 \Rightarrow a^2 = 4$$

Given the constraint $a < 0$, we find $a = -2$. Substituting this into $a^2 - c = 5$, we get $4 - c = 5$, which yields $c = -1$. The circle equation is $x^2 + y^2 - 2x - 4y - 1 = 0$.

### The Tangent Hunt

Completing the square for the circle equation:
$$(x^2 - 2x + 1) + (y^2 - 4y + 4) = 1 + 4 + 1 \Rightarrow (x-1)^2 + (y-2)^2 = 6$$

The center of the circle is $C(1, 2)$ and the radius is $R = \sqrt{6}$. We require a tangent perpendicular to $x + 2y = 0$ (slope $-1/2$), so the tangent must have a slope $m = 2$.

The equation of a tangent with slope $m$ is given by $y - y_c = m(x - x_c) \pm R\sqrt{1+m^2}$. Substituting our values:
$$y - 2 = 2(x - 1) \pm \sqrt{6}\sqrt{1 + 2^2}$$
$$y - 2 = 2x - 2 \pm \sqrt{30} \Rightarrow 2x - y \pm \sqrt{30} = 0$$

### Final Calculation

The shortest distance $d$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is calculated as:
$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

Applying this to the tangent lines $2x - y \pm \sqrt{30} = 0$:
$$d = \frac{|2(0) - 0 \pm \sqrt{30}|}{\sqrt{2^2 + (-1)^2}} = \frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$$

The shortest distance from the origin to the tangent is **$\sqrt{6}$**.

Select Answer:

Visualized Solution

Visual Anchor & Problem Setup

Logic Bridge: X-intercept Formula

Raw Setup: X-intercept Equation

Logic Bridge: Y-intercept Formula

Raw Setup: Y-intercept Equation

Atomic Compute: Solving for Constants

The Circle Equation & Center

Logic Bridge: Slope of the Tangent

Raw Setup: Equation of the Tangent

Atomic Compute: Simplifying Tangent Equation

The Way Forward: Distance from Origin

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Decoding the Circle

The Tangent Hunt

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visual Anchor & Problem Setup

Logic Bridge: X-intercept Formula

Raw Setup: X-intercept Equation

Logic Bridge: Y-intercept Formula

Raw Setup: Y-intercept Equation

Atomic Compute: Solving for Constants

The Circle Equation & Center

Logic Bridge: Slope of the Tangent

Raw Setup: Equation of the Tangent

Atomic Compute: Simplifying Tangent Equation

The Way Forward: Distance from Origin

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Decoding the Circle

The Tangent Hunt

Final Calculation

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