Question

Find the equation of a circle passing through the origin if the line pair, xy - 3x + 2y -6 = 0 is orthogonal to it. If this circle is orthogonal to the circle x 2 + y 2 − k x + 2 k y − 8 = 0 then find the value of k.

Answer 1

1

Answer 2

The problem involves two main parts: first, finding the equation of a circle based on given conditions, and second, using this circle to find the value of 'k' for another circle given an orthogonality condition.

Part 1: Finding the equation of the first circle

1. Analyze the line pair: The given line pair equation is $xy - 3x + 2y - 6 = 0$. We can factor this equation: $x(y-3) + 2(y-3) = 0$ $(x+2)(y-3) = 0$ This equation represents two straight lines: $L_1: x+2 = 0 \implies x = -2$ and $L_2: y-3 = 0 \implies y = 3$.

2. Interpret "line pair is orthogonal to the circle": In coordinate geometry, if a line is said to be "orthogonal" to a circle, it implies that the line passes through the center of the circle. This means the lines $x=-2$ and $y=3$ are diameters of the circle. The intersection of two diameters is the center of the circle. Therefore, the center of the first circle, let's call it $C_1$, is $(-2, 3)$.

3. Use the center and the origin to find the equation of $C_1$: The general equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the center $(-2, 3)$, we get: $(x - (-2))^2 + (y - 3)^2 = r^2$ $(x+2)^2 + (y-3)^2 = r^2$ The problem states that the circle passes through the origin $(0,0)$. Substitute $(0,0)$ into the equation to find $r^2$: $(0+2)^2 + (0-3)^2 = r^2$ $2^2 + (-3)^2 = r^2$ $4 + 9 = r^2$ $r^2 = 13$ So, the equation of the first circle is $(x+2)^2 + (y-3)^2 = 13$. Expanding this equation: $x^2 + 4x + 4 + y^2 - 6y + 9 = 13$ $x^2 + y^2 + 4x - 6y + 13 = 13$ $x^2 + y^2 + 4x - 6y = 0$. Comparing this with the general form $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$, we have: $2g_1 = 4 \implies g_1 = 2$ $2f_1 = -6 \implies f_1 = -3$ $c_1 = 0$

Part 2: Finding the value of k

1. Identify parameters of the second circle: The second circle is given by $C_2: x^2 + y^2 - kx + 2ky - 8 = 0$. Comparing this with the general form $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$, we have: $2g_2 = -k \implies g_2 = -k/2$ $2f_2 = 2k \implies f_2 = k$ $c_2 = -8$

2. Apply the condition for orthogonality of two circles: Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ are orthogonal if and only if: $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ Substitute the values we found for $C_1$ and $C_2$: $2(2)(-k/2) + 2(-3)(k) = 0 + (-8)$ $-2k - 6k = -8$ $-8k = -8$ $k = 1$

The value of k is 1.

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Factor the line pair equation $(x+2)(y-3)=0$ to identify the lines $x=-2$ and $y=3$.
2. Interpret "line pair is orthogonal to the circle" as the lines being diameters, meaning they pass through the circle's center. The intersection of these lines, $(-2,3)$, is the center of the first circle.
3. Use the center $(-2,3)$ and the fact that the circle passes through the origin $(0,0)$ to find the radius squared $r^2 = (0-(-2))^2 + (0-3)^2 = 4+9=13$.
4. Write the equation of the first circle as $x^2+y^2+4x-6y=0$. From this, identify $g_1=2, f_1=-3, c_1=0$.
5. From the second circle $x^2+y^2-kx+2ky-8=0$, identify $g_2=-k/2, f_2=k, c_2=-8$.
6. Apply the orthogonality condition for two circles, $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
7. Substitute the values: $2(2)(-k/2) + 2(-3)(k) = 0 + (-8)$, which simplifies to $-2k - 6k = -8$.
8. Solve for k: $-8k = -8 \implies k=1$.