Tangent L_1 equiv 3x - 4y - 8 = 0 and the cho | vedclass

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

Question

(D) Let the centre of the circle be $O(h, k)$ and radius be $r$. Since $L_1$ is a tangent,the perpendicular distance from $(h, k)$ to $3x - 4y - 8 = 0

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the centre of the circle be $O(h, k)$ and radius be $r$. Since $L_1$ is a tangent,the perpendicular distance from $(h, k)$ to $3x - 4y - 8 = 0$ is equal to $r$.$\frac{|3h - 4k - 8|}{\sqrt{3^2 + (-4)^2}} = r \Rightarrow |3h - 4k - 8| = 5r$. Since $L_1$ is a tangent,$r = 2$,so $|3h - 4k - 8| = 10$.Also,the distance from $(h, k)$ to $L_2 \equiv x + y - 1 = 0$ is $\sqrt{2}$.$\frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \sqrt{2} \Rightarrow |h + k - 1| = 2$.Given $h^2 + k^2 = 13$. Solving these equations,we find $(h, k) = (3, 2)$ or $(-2, 3)$.For $(h, k) = (3, 2)$,the midpoint $(\alpha, \beta)$ of the chord $L_2$ is the projection of the centre $(3, 2)$ on $x + y - 1 = 0$.$\frac{\alpha - 3}{1} = \frac{\beta - 2}{1} = -\frac{3 + 2 - 1}{1^2 + 1^2} = -\frac{4}{2} = -2$.$\alpha = 3 - 2 = 1, \beta = 2 - 2 = 0$. Thus,$\alpha + \beta = 1$.Given $r = 2$,we have $\alpha + \beta + r = 1 + 2 = 3$.

Similar Questions

The radius of the circle whose arc of length $15 \ cm$ makes an angle of $3/4$ radian at the centre is ..... $cm$.

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle in $Q$ with $PQ=5$,then $c$ is equal to

$A$ line drawn through the point $A(5,7)$ cuts the circle $x^2+y^2-36=0$ at the points $P$ and $Q$. Then,$AP \cdot AQ=$

If the radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $r$,then it will touch both the axes if:

The number of circles that touch all the three lines $x+y-1=0$,$x-y-1=0$,and $y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module