Question

Combined equation to the pair of tangents drawn from the origin to the circle $x^2 + y^2 + 4x + 6y + 9 = 0$ is

• A. 3(x^2 + y^2) = (x + 2y)^2
• B. 2(x^2 + y^2) = (3x + y)^2
• C. 9(x^2 + y^2) = (2x + 3y)^2
• D. x^2 + y^2 = (2x + 3y)^2

Answer 1

Answer: (C)

9(x^2 + y^2) = (2x + 3y)^2

Answer 2

The equation of the circle is $S = x^2 + y^2 + 4x + 6y + 9 = 0$. The general form of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given equation with the general form, we have: $2g = 4 \implies g = 2$ $2f = 6 \implies f = 3$ $c = 9$

The point from which the tangents are drawn is the origin, so $(x_1, y_1) = (0,0)$.

The combined equation of the pair of tangents from an external point $(x_1, y_1)$ to a circle $S=0$ is given by the formula $SS_1 = T^2$, where: $S = x^2 + y^2 + 2gx + 2fy + c$ $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$ $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$

Substitute $(x_1, y_1) = (0,0)$, $g=2$, $f=3$, and $c=9$: $S = x^2 + y^2 + 4x + 6y + 9$ $S_1 = (0)^2 + (0)^2 + 4(0) + 6(0) + 9 = 9$ $T = x(0) + y(0) + 2(x+0) + 3(y+0) + 9 = 2x + 3y + 9$

Now, apply the formula $SS_1 = T^2$: $(x^2 + y^2 + 4x + 6y + 9) \cdot 9 = (2x + 3y + 9)^2$

Let's simplify this equation. The equation $SS_1 = T^2$ is derived from the property that the equation of the pair of tangents from $(x_1, y_1)$ to $S=0$ is $S \cdot S(x_1, y_1) = (T(x,y))^2$.

The equation of the circle is $x^2 + y^2 + 4x + 6y + 9 = 0$. The point is $(0,0)$. $S = x^2 + y^2 + 4x + 6y + 9$. $S_1 = 0^2 + 0^2 + 4(0) + 6(0) + 9 = 9$. $T = x(0) + y(0) + 2(x+0) + 3(y+0) + 9 = 2x + 3y + 9$.

The combined equation of tangents is $S \cdot S_1 = T^2$. $(x^2 + y^2 + 4x + 6y + 9) \cdot 9 = (2x + 3y + 9)^2$.

Let's re-examine the formula for the combined equation of tangents from the origin to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$. The equation is $c(x^2+y^2) = (gx+fy)^2$. In this case, $g=2$, $f=3$, and $c=9$. So, the equation becomes: $9(x^2 + y^2) = (2x + 3y)^2$.

Let's verify this formula. The general equation of a pair of tangents from the origin to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $c(x^2+y^2) = (gx+fy)^2$. Here, $g=2$, $f=3$, and $c=9$. Substituting these values, we get: $9(x^2+y^2) = (2x+3y)^2$.

Expanding this: $9x^2 + 9y^2 = (2x)^2 + 2(2x)(3y) + (3y)^2$ $9x^2 + 9y^2 = 4x^2 + 12xy + 9y^2$ $9x^2 = 4x^2 + 12xy$ $5x^2 - 12xy = 0$ $x(5x - 12y) = 0$. This represents the two tangent lines $x=0$ and $5x-12y=0$.

Now, let's check the options. (A) $3(x^2 + y^2) = (x + 2y)^2 \implies 3x^2 + 3y^2 = x^2 + 4xy + 4y^2 \implies 2x^2 - 4xy - y^2 = 0$. (B) $2(x^2 + y^2) = (3x + y)^2 \implies 2x^2 + 2y^2 = 9x^2 + 6xy + y^2 \implies -7x^2 - 6xy + y^2 = 0$. (C) $9(x^2 + y^2) = (2x + 3y)^2 \implies 9x^2 + 9y^2 = 4x^2 + 12xy + 9y^2 \implies 5x^2 - 12xy = 0$. This matches. (D) $x^2 + y^2 = (2x + 3y)^2 \implies x^2 + y^2 = 4x^2 + 12xy + 9y^2 \implies -3x^2 - 12xy - 8y^2 = 0$.

Therefore, option (C) is the correct answer.