The tangent(s) from the point of intersection | vedclass

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

Question

(B) First,find the point of intersection of the lines $2x - 3y + 1 = 0$ and $3x - 2y - 1 = 0$. Solving these equations,we get $x = 1$ and $y = 1$. Thu

Vedclass · Accepted Answer

$2x - y - 1 = 0$

Vedclass · Answer

(B) First,find the point of intersection of the lines $2x - 3y + 1 = 0$ and $3x - 2y - 1 = 0$. Solving these equations,we get $x = 1$ and $y = 1$. Thus,the point of intersection is $(1, 1)$. Next,check if $(1, 1)$ lies on the circle $x^2 + y^2 + 2x - 4y = 0$. Substituting $(1, 1)$ into the circle equation: $(1)^2 + (1)^2 + 2(1) - 4(1) = 1 + 1 + 2 - 4 = 0$. Since the point $(1, 1)$ lies on the circle,the tangent at this point is given by the formula $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$. Here,$g = 1, f = -2, c = 0, x_1 = 1, y_1 = 1$. Substituting these values: $x(1) + y(1) + 1(x + 1) - 2(y + 1) + 0 = 0$. $x + y + x + 1 - 2y - 2 = 0$. $2x - y - 1 = 0$.

The tangent$(s)$ from the point of intersection of the lines $2x - 3y + 1 = 0$ and $3x - 2y - 1 = 0$ to the circle $x^2 + y^2 + 2x - 4y = 0$ will be -

Similar Questions

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at $(1,-2)$ is

If $P(-9,-1)$ is a point on the circle $x^2+y^2+4x+8y-38=0$,then find the equation of the tangent drawn at the other end of the diameter drawn through $P$.

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 $........$.

The lines $y - y_1 = m(x - x_1) \pm a \sqrt{1 + m^2}$ are tangents to the same circle. The radius of the circle is:

If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2}+y^{2}=4$ from the point $(3,2)$,then $m_{1}-m_{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module