Question

If the angle between the pair of tangents drawn to the circle x2 + y2 - 2x + 4y + 3 = 0 from the point (6, -5) is

• A. $7x^2+23y^2+30xy+66x + 50y−73=0$
• B. $7x^2+23y^2−30xy+66x−50y−73=0$
• C. $7x^2+3y^2+30xy−66x+50y−73=0$
• D. $3x^2+7y^2+30xy+66x+50y−73=0$

Answer 1

Answer: (A)

$7x^2+23y^2+30xy+66x + 50y−73=0$

Answer 2

The given circle equation, $x^2+y^2−2x+4y+3=0$, can be rewritten as $(x−1)^2+(y+2)^2=2$.

Let's consider the equation of the line as $y=mx+c$. Since it passes through the point (6, -5), we can express $c=−(5+6m)$. So, the equation of the line becomes $y=mx−(5+6m)$.

By substituting this line equation into the circle equation, $(x−1)^2+(mx+2−(5+6m))2=0$, and upon expanding and rearranging the terms, we get
$(1+m^2)x^2−x(12m^2+6m+2)+(36m^2+36m+8)=0$.

This equation has a unique solution because the line is a tangent to the given circle. Utilizing the quadratic formula $b^2=4ac$, we have $(12m^2+6m+2)2=4(1+m2)(36m^2+36m+8)$.

Rearranging the terms leads to the equation
$23m^2+30m+7=0$, marked as (1).

Now, let the line equations be $y=m_1​x+c_1$​ and $y=m_2​x+c_2​$, where:
$c_1​=−(5+6m_1​)$ and $c_2​=−(5+6m_2​)$.

The equation of the pair of lines is $(m_1​x−y+c_1​)(m_2​x−y+c_2​)=0$.

Expanding this, we get
$m_1​m_2​x^2+y^2+(m_1​+m_2​)xy+(m_1​c_2​+c_1​m_2​)x−y(c_1​+c_2​)+c_1​c_2​=0$.

From equation (1),
$m_1​+m_2​=−\frac{23}{30}$​ and $m_1​m_2​=\frac{23}{7}$​.

$c_1​+c_2​=−(10+12(m_1​+m_2​))=\frac{23}{-50}​$.
$c_1​m_2​+m_1​c_2​=−(5(m_1​+m_2​)+12m_1​m_2​)=\frac{23}{66}​$.
$c_1​c_2​=(5+6m_1​)(5+6m_2​)=\frac{23}{−73}​$.

Substituting the corresponding values, we get the equation $7x^2+23y^2+30xy+66x+50y−73=0$.