If the circle C1 : x² + y² = 16 intersects another circle C2 of radius 5 in such a manner that commo

Q: Wait, why do we assume the common chord is the diameter of $C_1$? Couldn't it be the diameter of $C_2$?

That is a fantastic question! Remember, the common chord must exist in *both* circles. If it were the diameter of $C_2$ (which has a radius of 5), the chord length would be 10. However, the maximum chord length possible in $C_1$ (radius 4) is only 8. A chord of length 10 simply cannot fit inside $C_1$. Therefore, the common chord must be the diameter of the smaller circle, $C_1$.

Q: Why is the line of centers perpendicular to the common chord?

This is a classic property of intersecting circles! The common chord is a chord for both circles. The perpendicular bisector of any chord always passes through the center of that circle. Since the common chord is a chord for *both* $C_1$ and $C_2$, the line of centers must be the perpendicular bisector for both. Hence, the line of centers is perpendicular to the common chord.

Q: Why are there two possible answers for the center of $C_2$?

Geometry is symmetric! The problem tells us the slope of the chord, but it doesn't specify which side of the chord the center of $C_2$ lies on. The center $O_2$ could be 'above' the chord or 'below' it, while still maintaining the same distance and perpendicularity. Both positions are geometrically valid, which is why we get two coordinate pairs.

Question

If the circle $C_1 : x^2 + y^2 = 16$ intersects another circle $C_2$ of radius 5 in such a manner that common chord is of maximum length and has a slope equal to $3/4$, then the coordinates of the centre of $C_2$ are .........

Sigma Percentile · Accepted Answer

### The Dance of Two Circles: A Geometric Journey

My dear student, welcome to the beautiful world of coordinate geometry. Today, we are not just solving for coordinates; we are uncovering the hidden symmetry between two circles.

Imagine you are standing on a plane, and you see two circles, $C_1$ and $C_2$, locked in an embrace, intersecting at two points. The problem asks us to find the center of $C_2$, given that their common chord is of maximum length.

### Phase 1: The Revelation of the Maximum Chord

First, look at $C_1: x^2 + y^2 = 16$. This is a circle centered at the origin $(0, 0)$ with a radius $r_1 = 4$.

Now, we are told the common chord is of 'maximum length.' In any circle, the longest chord is the diameter.

Since $C_1$ has a radius of $4$ and $C_2$ has a radius of $5$, $C_1$ is the smaller circle. If the common chord were the diameter of $C_2$, it would have a length of $10$, which is impossible to fit inside $C_1$.

Thus, the common chord **must** be the diameter of $C_1$. This means the chord passes through the origin $(0, 0)$ and has a length of $8$.

### Phase 2: The Geometry of Centers

We know the slope of this chord is $m = \frac{3}{4}$. Since it passes through the origin, its equation is simply $y = \frac{3}{4}x$, or $3x - 4y = 0$.

Now, here is the JEE-favorite concept: the line joining the centers of two intersecting circles is always perpendicular to their common chord.

If the slope of our chord is $m_1 = \frac{3}{4}$, then the slope of the line connecting the centers, $O_1O_2$, must be the negative reciprocal: 
$$m_2 = -\frac{4}{3}$$

### Phase 3: The Right Triangle

Let us visualize the triangle formed by the center of the first circle $O_1$, the center of the second circle $O_2$, and one of the intersection points $A$. We know $O_1A = r_1 = 4$ and $O_2A = r_2 = 5$.

Because the common chord is the diameter of $C_1$, the angle at $O_1$ is $90^\circ$. We have a right-angled triangle $	riangle O_1AO_2$.

By the Pythagorean theorem, the distance between the centers is:
$$O_1O_2 = \sqrt{r_2^2 - r_1^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3$$

### Phase 4: The Parametric Leap

We need the coordinates of $O_2$. We know it lies on a line with slope $-\frac{4}{3}$ at a distance of $3$ units from the origin.

Using the parametric form of a line, where $	an 	heta = -\frac{4}{3}$, we find $\cos 	heta = \pm \frac{3}{5}$ and $\sin 	heta = \mp \frac{4}{5}$. The coordinates are given by $(x_1 \pm d \cos 	heta, y_1 \pm d \sin 	heta)$.

Substituting our values: 
$$h = 0 \pm 3\left(\frac{3}{5}ight) = \pm \frac{9}{5}$$ 
$$k = 0 \mp 3\left(\frac{4}{5}ight) = \mp \frac{12}{5}$$

This gives us our two possible centers: **$(\frac{9}{5}, -\frac{12}{5})$ and $(-\frac{9}{5}, \frac{12}{5})$**. You have successfully navigated the geometry!

Animated Solution for Mathematics - Circles: If the circle $C_{1} : x^{2} + y^{2} = 16$ intersects another circle $C_{2}$ of radius 5 in such a manner that common chord is of maximum length and has a slope equal to $3/4$ , then the coordinates of the centre of $C_{2}$ are .........

Visualized Solution

Analyze Circle $C_{1}$

The Maximum Chord Condition

Equation of the Common Chord

Geometry of the Centers

The Right Triangle

Distance Between Centers

Parametric Coordinates of $O_{2}$

Calculating Coordinates

Final Conclusion

The Sigma Insight: Length of Tangent and Chord of Contact

The Dance of Two Circles

Phase 1

Phase 2

Phase 3

Phase 4

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