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  • Tangents are drawn to a unit circle with centre at the origin from each point on the line <img alt="\mathrm{2 x+y=4}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2%20x&plus

Tangents are drawn to a unit circle with centre at the origin from each point on the line \mathrm{2 x+y=4}. Then the equation to the locus of the middle point of the chord of contact is

Option: 1

2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{x}+\mathrm{y}


Option: 2

2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{x}+2 \mathrm{y}


Option: 3

\mathrm{4\left(x^{2}+y^{2}\right)=2 x+y}


Option: 4

none


Answers (1)

best_answer

\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ lies on $2 \mathrm{x}+\mathrm{y}=4

\Rightarrow \quad 2 \mathrm{x}_{1}+\mathrm{y}_{1}=4 \quad \cdots(1)
chord of contact w.r.t. \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)
\mathrm{xx}_{1}+\mathrm{yy}_{1}=1
also equation of chord whose mid point is (\mathrm{h}, \mathrm{k})
\mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{hx}+\mathrm{ky}

\mathrm{\therefore \quad \frac{\mathrm{x}_{1}}{\mathrm{~h}}=\frac{\mathrm{y}_{1}}{\mathrm{k}}=\frac{1}{\mathrm{~h}^{2}+\mathrm{k}^{2}} \quad \Rightarrow \quad \mathrm{x}_{1}=\frac{\mathrm{h}}{\mathrm{h}^{2}+\mathrm{k}^{2}} ; \mathrm{y}_{1}=\frac{\mathrm{k}}{\mathrm{h}^{2}+\mathrm{k}^{2}}}

substitute in (1)
\mathrm{2 \cdot \frac{\mathrm{h}}{\mathrm{h}^{2}+\mathrm{k}^{2}}+\frac{\mathrm{k}}{\mathrm{h}^{2}+\mathrm{k}^{2}}=4}

\mathrm{\text { locus } =4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=2 \mathrm{x}+\mathrm{y} \quad}

Posted by

Ritika Jonwal

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