The equation of any tangent to the circle $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ is

y = m (x - 1) + 3 \sqrt{1 + m^{2}} - 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y = m x + 3 \sqrt{1 + m^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y = m x + 3 \sqrt{1 + m^{2}} - 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $y = m (x - 1) + 3 \sqrt{1 + m^{2}} - 2$
Centre and radius of the given circle is $C (1, - 2)$ and $\sqrt{1 + 4 + 4} = 3$ respectively.
Let $y = m x + c$ be the tangent to the circle. Thus distance of line from centre of the circle and radius will be equal.
$\Rightarrow ∣ ∣ ∣ \frac{m + c + 2}{\sqrt{1 + m^{2}}} ∣ ∣ ∣ = r = 3 \Rightarrow c = \pm 3 \sqrt{1 + m^{2}} - (m + 2)$
Hence required tangent is , $y = m x \pm 3 \sqrt{1 + m^{2}} - (m + 2)$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 4 y = 0,

x^{3} - y^{2} = 0

(m^{2}, - m^{3})

= - \frac{1}{m} x - 2 m^{3},

m^{2}

\begin{matrix} C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) x^{2} + y^{2} = a^{2} & (i) m y = m^{2} x + a & (P) (\frac{a}{m^{2}}, \frac{2 a}{m}) (I I) x^{2} + a^{2} y^{2} = a^{2} & (i i) y = m x + a \sqrt{m^{2} + 1} & (Q) (\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}}) (I I I) y^{2} = 4 a x & (i i i) y = m x + \sqrt{a^{2} m^{2} - 1} & (R) (\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}}) (I V) x^{2} - a^{2} y^{2} = a^{2} & (i v) y = m x + \sqrt{a^{2} m^{2} + 1} & (S) (\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}}) \end{matrix}

a = \sqrt{2}

(- 1, 1)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 ?

If the line $m (x - 1) - m^{2} (y - 1) + 1 = 0$ is a tangent to a parabola for any real value of m, then the equation of the parabola is

Join BYJU'S Learning Program