The area of the triangle formed by the tangen | vedclass

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

Question

(C) The equation of the tangent at $P(3, 4)$ to the circle ${x^2} + {y^2} = 25$ is given by $xx_1 + yy_1 = r^2$,which is $3x + 4y = 25$.To find the in

Vedclass · Accepted Answer

$\frac{625}{24}$

Vedclass · Answer

(C) The equation of the tangent at $P(3, 4)$ to the circle ${x^2} + {y^2} = 25$ is given by $xx_1 + yy_1 = r^2$,which is $3x + 4y = 25$.To find the intercepts on the coordinate axes,set $y = 0$ to get $3x = 25$,so $x = \frac{25}{3}$. Thus,the point $A$ is $(\frac{25}{3}, 0)$.Set $x = 0$ to get $4y = 25$,so $y = \frac{25}{4}$. Thus,the point $B$ is $(0, \frac{25}{4})$.The triangle formed by the tangent and the coordinate axes is a right-angled triangle $\Delta OAB$ with base $OA = \frac{25}{3}$ and height $OB = \frac{25}{4}$.Area of $\Delta OAB = \frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes \frac{25}{3} 	imes \frac{25}{4} = \frac{625}{24}$.

The area of the triangle formed by the tangent at $(3, 4)$ to the circle ${x^2} + {y^2} = 25$ and the coordinate axes is

Similar Questions

If the line $y=2x+c$ is a tangent to the circle $x^2+y^2=5$,then a value of $c$ is

If the tangent at a point $P$ on the circle $x^2 + y^2 + 6x + 6y = 2$ meets the line $5x - 2y + 6 = 0$ at a point $Q$ on the $y$-axis,then the length of $PQ$ is . . . . .

Two tangents drawn from $P(1, 7)$ to the circle $x^2 + y^2 = 25$ touch the circle at $Q$ and $R$ respectively. The area of the quadrilateral $PQOR$ is

The slope of a common tangent to the circles $x^2+y^2-4x-8y+16=0$ and $x^2+y^2-6x-16y+64=0$ is

$A$ tangent is drawn to the circle $2x^{2} + 2y^{2} - 3x + 4y = 0$ at point $A$ and it meets the line $x + y = 3$ at $B(2, 1)$. Then,the length of $AB$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module