Question

Equation as a common tangent with positive slope to the circle as well as to the hyperbola is
A. \[2x - \sqrt {5y} - 20 = 0\]
B. \[2x - \sqrt {5y} + 4 = 0\]
C. \[3x - 4y + 8 = 0\]
D. \[4x - 3y + 4 = 0\]

Answer 1

Hint: Tangent to a circle is the line that touches the circle at only one point. There can be only one tangent at a point to circle. Point of tangency is the point at which tangent meets the circle. Transverse tangents or Direct common tangents always meet on the line joining centres of the two circles.

Complete step by step solution:
As we know the standard equation of Hyperbola is given by:
\[\dfrac{x}{{{a^2}}} - \dfrac{y}{{{b^2}}} = 1\]
Let the tangent to Hyperbola:
\[\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1\]
Is \[y = mx + \sqrt {9{m^2} - 4m} > 0\]
Let the tangent to circle:
\[{x^2} + {y^2} - 8x = 0\]
\[\dfrac{{4m + \sqrt {9{m^2} - 4} }}{{\sqrt {1 + {m^2}} }} = 4\]
Equating the terms, we get:
\[495{m^4} + 104{m^2} - 400 = 0\]
\[{m^2} = \dfrac{4}{5}\]
\[m = \dfrac{2}{{\sqrt 5 }}\]
Therefore, the tangent is
\[y = \dfrac{2}{{\sqrt 5 }}m + \dfrac{4}{{\sqrt 5 }}\]
Hence, equating the terms we get
\[2x - \sqrt {5y} + 4 = 0\]

Hence, option B is the right answer.

Additional information:
A line that touches the circle at a single point is known as a tangent to a circle. The point where tangent meets the circle is called the point of tangency. The tangent is perpendicular to the radius of the circle, with which it intersects. Tangent can be considered for any curved shapes. Since tangent is a line, hence it also has its equation.
Properties of Tangent are:
The tangent always touches the circle at a single point.
It is perpendicular to the radius of the circle at the point of tangency.
It never intersects the circle at two points.
The length of tangents from an external point to a circle are equal.

Note: A differential equation is homogeneous if it is a similar function of the anonymous function and its derivatives. For linear differential equations, there are no constant terms. The solutions of any linear ordinary differential equation of any degree or order may be calculated by integration from the solution of the homogeneous equation achieved by eliminating the constant term.