Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The tangents drawn from the origin to the circle  <img alt="\mathrm{x^2+y^2-2 r x-2 h y+h^2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-2%20r

The tangents drawn from the origin to the circle  \mathrm{x^2+y^2-2 r x-2 h y+h^2=0} are perpendicular if

Option: 1

h=r


Option: 2

h=-r


Option: 3

\mathrm{\mathrm{r}^2+\mathrm{h}^2=1}


Option: 4

\mathrm{\mathrm{r}^2=\mathrm{h}^2}


Answers (1)

best_answer

The combined equation of the tangents drawn from (0,0), to

\mathrm{\begin{aligned} & \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{rx}-2 \mathrm{hy}+\mathrm{h}^2=0 \text { is } \\\\ & \left(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{rx}-2 \mathrm{hy}+\mathrm{h}^2\right) \mathrm{h}^2=\left(-\mathrm{rx}-\mathrm{hy}+\mathrm{h}^2\right)^2 \end{aligned}}

This equation represents a pair of perpendicular straight lines if coeff. of \mathrm{x^2+\text { coeff. of } y^2=0} i.e.

\mathrm{\begin{array}{ll} & 2 h^2-r^2-h^2=0 \\\\ \Rightarrow \quad & r^2=h^2 \text { or } r= \pm h \end{array}}

Hence (a), (b) and (d) are correct answers.

Posted by

Anam Khan

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?
anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?
anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question