Let O be the centre of the circle x ^2+ y ^2= | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(image)

$M - I$

$OA =\frac{\sqrt{5}}{2} \quad OC =\frac{4}{\sqrt{5}}$

$CQ = OC =\frac{4}{\sqrt{5}} 	ext { and } CA =\frac{3}{2 \sqrt{5}}$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(image)

$M - I$

$OA =\frac{\sqrt{5}}{2} \quad OC =\frac{4}{\sqrt{5}}$

$CQ = OC =\frac{4}{\sqrt{5}} 	ext { and } CA =\frac{3}{2 \sqrt{5}}$

$	herefore \quad O Q=\sqrt{ OA ^2+ AQ ^2}=\sqrt{ OA ^2+\left( CQ ^2- CA ^2ight)}$

$\Rightarrow \quad \sqrt{\frac{5}{4}+\frac{16}{5}-\frac{9}{20}}=\sqrt{4}$

$\Rightarrow \quad 2= r$

(image)

$P Q: h x+k y=r^2$

$	ext { Given } P Q \quad 2 x+4 y=5$

$\Rightarrow \quad \frac{h}{2}=\frac{k}{4}=\frac{r^2}{5} \Rightarrow h=\frac{2 r^2}{5}\quad k=\frac{4 r^2}{5}$

$	herefore \quad C=\left(\frac{r^2}{5}, \frac{2 r^2}{5}ight)$

$	herefore \quad C 	ext { lies on } x+2 y=4 \quad \Rightarrow \quad \frac{r^2}{5}+2\left(\frac{2 r^2}{5}ight)=4$

$\Rightarrow \quad r^2=4 \Rightarrow r=2$

Similar Questions

The line $lx + my + n = 0$ will be a tangent to the circle ${x^2} + {y^2} = {a^2}$ if

If $\theta $ is the angle subtended at $P({x_1},{y_1})$ by the circle $S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$, then

The equations of the tangents drawn from the point $(0, 1)$ to the circle ${x^2} + {y^2} - 2x + 4y = 0$ are

The line $y = x + c$will intersect the circle ${x^2} + {y^2} = 1$ in two coincident points, if

An infinite number of tangents can be drawn from $(1, 2)$ to the circle ${x^2} + {y^2} - 2x - 4y + \lambda = 0$, then $\lambda = $