3x + 4y - 43 = 0 is a tangent to the circle S | vedclass

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

Question

(C) The equation of the circle is $x^2 + y^2 - 6x + 8y + k = 0$. The centre $C$ is $(3, -4)$ and the radius $r$ is $\sqrt{3^2 + (-4)^2 - k} = \sqrt{25

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) The equation of the circle is $x^2 + y^2 - 6x + 8y + k = 0$. The centre $C$ is $(3, -4)$ and the radius $r$ is $\sqrt{3^2 + (-4)^2 - k} = \sqrt{25 - k}$.Since $3x + 4y - 43 = 0$ is a tangent at $P$,the distance from $C(3, -4)$ to the line is equal to the radius $r$.$r = \frac{|3(3) + 4(-4) - 43|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 43|}{5} = \frac{|-50|}{5} = 10$.Thus,$r^2 = 100$,so $25 - k = 100$,which gives $k = -75$.The circle is $S \equiv x^2 + y^2 - 6x + 8y - 75 = 0$.Point $Q$ divides $CP$ in the ratio $-1:2$. Let $C = (3, -4)$ and $P = (x_p, y_p)$.$Q = \frac{-1(P) + 2(C)}{-1 + 2} = 2C - P = 2(3, -4) - P = (6 - x_p, -8 - y_p)$.The power of point $Q$ with respect to the circle $S=0$ is $S(Q) = x_Q^2 + y_Q^2 - 6x_Q + 8y_Q - 75$.Since $P$ lies on the circle,$x_p^2 + y_p^2 - 6x_p + 8y_p - 75 = 0$.Substituting $Q$ into the circle equation: $(6-x_p)^2 + (-8-y_p)^2 - 6(6-x_p) + 8(-8-y_p) - 75 = 0$.$36 - 12x_p + x_p^2 + 64 + 16y_p + y_p^2 - 36 + 6x_p - 64 - 8y_p - 75 = (x_p^2 + y_p^2 - 6x_p + 8y_p - 75) = 0$ is not correct here. Actually,the power of point $Q$ is $CQ^2 - r^2$. Since $Q$ divides $CP$ in ratio $-1:2$,$CQ = |-1| 	imes CP / |2-1| = r$. Thus $CQ^2 = r^2$,so the power of point $Q$ is $r^2 - r^2 = 0$.

$3x + 4y - 43 = 0$ is a tangent to the circle $S \equiv x^2 + y^2 - 6x + 8y + k = 0$ at a point $P$. If $C$ is the centre of the circle and $Q$ is a point which divides $CP$ in the ratio $-1:2$,then the power of the point $Q$ with respect to the circle $S = 0$ is

Similar Questions

$O(0,0)$ and $A(1,0)$ are the centers of two unit circles $C_1$ and $C_2$ respectively. $C_3$ is also a unit circle having its center above the $X$-axis and passing through $O$ and $A$. The equation of the common tangent to $C_1$ and $C_3$ which does not intersect the circle $C_2$ is

$A$ circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is

Two sides of a square are along the lines $x=-5$ and $y=4$. The point of intersection of the diagonals is $(3,-4)$. The point of intersection of the tangents drawn to the circumcircle of the square at the two consecutive vertices lying on $x=-5$ is

Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at the point $P(-4,0)$. Then the area of the quadrilateral $PAOB$,where $O$ is the origin,is:

If the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ cuts each of the three circles $x^2 + y^2 + 4x + 4y + 7 = 0$,$x^2 + y^2 - 4x + 4y + 7 = 0$,and $x^2 + y^2 - 4x - 4y + 7 = 0$ orthogonally,then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $S = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module