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

The Correct Option is A

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 \(7x^2+23y^2+30xy+66x+50y−73=0\).