Let the centre of a circle C be (alpha, beta) | vedclass

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

Question

(A) The centre $(\alpha, \beta)$ lies on the normal line $4x + 3y = 1$,so $4\alpha + 3\beta = 1$,which gives $\beta = \frac{1 - 4\alpha}{3}$.The dista

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) The centre $(\alpha, \beta)$ lies on the normal line $4x + 3y = 1$,so $4\alpha + 3\beta = 1$,which gives $\beta = \frac{1 - 4\alpha}{3}$.The distance from the centre $(\alpha, \beta)$ to the two tangents $3x + 4y - 24 = 0$ and $3x - 4y - 32 = 0$ must be equal to the radius $r$.Thus,$r = \left| \frac{3\alpha + 4\beta - 24}{5} \right| = \left| \frac{3\alpha - 4\beta - 32}{5} \right|$.Substituting $\beta = \frac{1 - 4\alpha}{3}$:$r = \left| \frac{3\alpha + 4(\frac{1 - 4\alpha}{3}) - 24}{5} \right| = \left| \frac{9\alpha + 4 - 16\alpha - 72}{15} \right| = \left| \frac{-7\alpha - 68}{15} \right|$.$r = \left| \frac{3\alpha - 4(\frac{1 - 4\alpha}{3}) - 32}{5} \right| = \left| \frac{9\alpha - 4 + 16\alpha - 96}{15} \right| = \left| \frac{25\alpha - 100}{15} \right| = \left| \frac{5(\alpha - 4)}{3} \right|$.Equating the two expressions for $r$:$\left| \frac{-7\alpha - 68}{15} \right| = \left| \frac{25\alpha - 100}{15} \right|$ $\Rightarrow |-7\alpha - 68| = |25\alpha - 100|$.Case $1$: $-7\alpha - 68 = 25\alpha - 100$ $\Rightarrow 32 = 32\alpha$ $\Rightarrow \alpha = 1$.Then $\beta = \frac{1 - 4(1)}{3} = -1$. Radius $r = \left| \frac{25(1) - 100}{15} \right| = \left| \frac{-75}{15} \right| = 5$. Since $r Case $2$: $-7\alpha - 68 = -(25\alpha - 100)$ $\Rightarrow -7\alpha - 68 = -25\alpha + 100$ $\Rightarrow 18\alpha = 168$ $\Rightarrow \alpha = \frac{28}{3}$.Then $\beta = \frac{1 - 4(28/3)}{3} = \frac{3 - 112}{9} = -\frac{109}{9}$. Radius $r = \left| \frac{25(28/3) - 100}{15} \right| = \left| \frac{700/3 - 300/3}{15} \right| = \frac{400}{45} = \frac{80}{9} \approx 8.88$. Since $r > 8$,this is rejected.Thus,$\alpha = 1, \beta = -1, r = 5$.$\alpha - \beta + r = 1 - (-1) + 5 = 7$.

Let the centre of a circle $C$ be $(\alpha, \beta)$ and its radius $r < 8$. Let $3x + 4y = 24$ and $3x - 4y = 32$ be two tangents and $4x + 3y = 1$ be a normal to $C$. Then $(\alpha - \beta + r)$ is equal to $........$.

Similar Questions

If the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ at $(\alpha, \beta)$,then $\alpha + 2\beta =$

If the pair of tangents drawn to the circle $x^2+y^2=a^2$ from the point $(10,4)$ are perpendicular,then $a=$

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

The length of the tangent drawn to the circle $x^2+y^2-2x+4y-11=0$ from the point $(1,3)$ is:

The square of the length of the tangent from $(3, -4)$ to the circle $x^2 + y^2 - 4x - 6y + 3 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module