Circle Ques 116
- The equations of the tangents drawn from the origin to the circle $x^{2}+y^{2}+2 r x+2 h y+h^{2}=0$, are
$(1988,2 M)$
(a) $x=0$
(b) $y=0$
(c) $\left(h^{2}-r^{2}\right) x-2 r h y=0$
(d) $\left(h^{2}-r^{2}\right) x+2 r h y=0$
Assertion and Reason
For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I.
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
Show Answer
Answer:
Correct Answer: 116.$(a, c)$
Solution:
Formula:
- Since, tangents are drawn from origin. So, the equation of tangent be $y=m x$.
$\Rightarrow$ Length of perpendicular from origin $=$ radius
$$ \begin{aligned} & \Rightarrow & \frac{|m r+h|}{\sqrt{m^{2}+1}} & =r \\ \Rightarrow & & m^{2} r^{2}+h^{2}+2 m r h & =r^{2}\left(m^{2}+1\right) \\ & \Rightarrow & m & =\frac{r^{2}-h^{2}}{2 r h}, m=\infty \end{aligned} $$
$\therefore$ Equation of tangents are $y=\frac{r^{2}-h^{2}}{2 r h} x, x=0$
Therefore (a) and (c) are the correct answers.
BETA
