MHT CET · Maths · Circle
The equation of tangents to the circle \(x^2+y^2=4\) which are parallel to \(x+2 y+3=0\) are
- A \(x+2 y= \pm 2 \sqrt{5}\)
- B \(x+2 y= \pm 2 \sqrt{3}\)
- C \(x-2 y= \pm 2\)
- D \(x-2 y= \pm 2 \sqrt{5}\)
Answer & Solution
Correct Answer
(A) \(x+2 y= \pm 2 \sqrt{5}\)
Step-by-step Solution
Detailed explanation
Equation of tangent to the circle \(x^2+y^2=r^2\) having slope \(m\) is
\(\Rightarrow y=-\frac{1}{2} x \pm 2 \sqrt{1+\left(\frac{-1}{2}\right)^2} \quad\)\(\left[\text { Here } \mathrm{r}=2 \text { and } \mathrm{m}=\frac{-1}{2}\right] \)
\( \Rightarrow \mathrm{y}=\frac{-\mathrm{x}}{2} \pm 2 \times \frac{\sqrt{5}}{2} \)
\( \Rightarrow 2 \mathrm{y}=-\mathrm{x} \pm 2 \sqrt{5} \)
\( \Rightarrow \mathrm{x}+2 \mathrm{y}= \pm 2 \sqrt{5}\)
\(\Rightarrow y=-\frac{1}{2} x \pm 2 \sqrt{1+\left(\frac{-1}{2}\right)^2} \quad\)\(\left[\text { Here } \mathrm{r}=2 \text { and } \mathrm{m}=\frac{-1}{2}\right] \)
\( \Rightarrow \mathrm{y}=\frac{-\mathrm{x}}{2} \pm 2 \times \frac{\sqrt{5}}{2} \)
\( \Rightarrow 2 \mathrm{y}=-\mathrm{x} \pm 2 \sqrt{5} \)
\( \Rightarrow \mathrm{x}+2 \mathrm{y}= \pm 2 \sqrt{5}\)
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