Find the equation of the circle whose radius is 5 and which touches the circle x² + y² - 2x - 4y - 2

Q: Wait, why can't we just use the family of circles equation $S + \lambda L = 0$ to solve this?

You absolutely could! That is a powerful tool in your arsenal. However, in JEE Advanced, time is your most precious currency. The geometric approach—recognizing the midpoint property—is significantly faster and reduces the chance of algebraic errors. Always look for the geometric shortcut before diving into heavy algebra.

Q: How do I know for sure that the circles touch externally and not internally?

Great question! If they touched internally, the distance between centers would be $|r_1 - r_2|$. Since $r_1 = 5$ and $r_2 = 5$, the distance would be zero, meaning the circles would be identical. Since the problem asks for a new circle, they must be externally tangent, where the distance between centers is $r_1 + r_2 = 10$.

Q: Is it possible there is more than one answer?

In this specific configuration, no. Once you fix the center of the first circle and the point of contact, the line connecting the centers is fixed. Since the radius is also fixed, the position of the second center is uniquely determined. You are on the right track!

Question

Find the equation of the circle whose radius is 5 and which touches the circle $x^2 + y^2 - 2x - 4y - 20 = 0$ at the point $(5, 5)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every great journey begins with understanding your starting point. We are given the equation $x^2 + y^2 - 2x - 4y - 20 = 0$.

To unlock its secrets, we compare it to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. By matching the coefficients, we find $2g = -2$ (which gives $g = -1$) and $2f = -4$ (which gives $f = -2$).

The center $C_1$ is defined as $(-g, -f)$, so our center is $(1, 2)$.

Now, for the radius, we use the formula $r_1 = \sqrt{g^2 + f^2 - c}$. Plugging in our values:

$$r_1 = \sqrt{(-1)^2 + (-2)^2 - (-20)} = \sqrt{1 + 4 + 20} = \sqrt{25} = 5$$

We have our first circle: centered at $(1, 2)$ with a radius of $5$.

### The Geometric Insight

The problem states that a new circle touches our first one at the point $P(5, 5)$. When two circles touch, they share a common tangent at the point of contact.

The centers of both circles and the point of contact $P$ must lie on a single straight line. This is a fundamental property of tangency.

Imagine a line passing through $C_1(1, 2)$ and $P(5, 5)$. The second circle, $C_2$, must lie on this line.

### The Midpoint Magic

We are told the new circle also has a radius of $5$. Note that the distance from $C_1$ to $P$ is $5$, and the distance from $P$ to $C_2$ is also $5$.

Because the distances are equal, the point of contact $P$ is the exact midpoint of the segment connecting $C_1$ and $C_2$. This is the "Aha!" moment.

Let the center of the new circle be $C_2(\alpha, \beta)$. Since $P(5, 5)$ is the midpoint of $C_1(1, 2)$ and $C_2(\alpha, \beta)$, we apply the midpoint formula:

$$\frac{1 + \alpha}{2} = 5 \quad 	ext{and} \quad \frac{2 + \beta}{2} = 5$$

Solving these is straightforward: $1 + \alpha = 10 \Rightarrow \alpha = 9$, and $2 + \beta = 10 \Rightarrow \beta = 8$. Our new center is $C_2(9, 8)$.

### Final Construction

We have arrived at the finish line with center $(9, 8)$ and radius $r_2 = 5$. The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.

Substituting our values, we get:

$$(x - 9)^2 + (y - 8)^2 = 5^2$$

Expanding this, we obtain:

$$(x^2 - 18x + 81) + (y^2 - 16y + 64) = 25$$

Combining the constants, we arrive at $x^2 + y^2 - 18x - 16y + 145 = 25$. This simplifies beautifully to the final equation:

**$x^2 + y^2 - 18x - 16y + 120 = 0$**

Animated Solution for Mathematics - Circles: Find the equation of the circle whose radius is 5 and which touches the circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at the point $(5, 5)$ .

Visualized Solution

The Given Circle

Center and Radius Formula

Finding the Center $C_{1}$

Calculating the Radius $r_{1}$

The Point of Contact $P$

The Required Circle $C_{2}$

The Midpoint Concept

Setting up the Midpoint Formula

Solving for $α$

Solving for $β$

Equation of the New Circle

Substituting the Values

Expanding the Equation

The Final Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Insight

The Midpoint Magic

Final Construction

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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 equation of the circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 at the point (5,5).

Visualized Solution

The Given Circle

Center and Radius Formula

Finding the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Calculating the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​

The Point of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Required Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​