The straight line x + 2y = 1 meets the coordi | vedclass

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

Question

(A) The line $x + 2y = 1$ intersects the $x$-axis at $A(1, 0)$ and the $y$-axis at $B(0, 1/2)$.Since the circle passes through the origin $(0, 0)$,$A(

Vedclass · Accepted Answer

$\frac{\sqrt{5}}{2}$

Vedclass · Answer

(A) The line $x + 2y = 1$ intersects the $x$-axis at $A(1, 0)$ and the $y$-axis at $B(0, 1/2)$.Since the circle passes through the origin $(0, 0)$,$A(1, 0)$,and $B(0, 1/2)$,and the triangle $OAB$ is a right-angled triangle at the origin,the segment $AB$ is the diameter of the circle.The equation of the circle with diameter $AB$ is $(x - 0)(x - 1) + (y - 0)(y - 1/2) = 0$,which simplifies to $x^2 + y^2 - x - \frac{1}{2}y = 0$.The tangent to the circle at the origin $(0, 0)$ is found by replacing $x^2$ with $x \cdot 0$,$y^2$ with $y \cdot 0$,$x$ with $\frac{x+0}{2}$,and $y$ with $\frac{y+0}{2}$.This gives $0 + 0 - \frac{x}{2} - \frac{1}{2} \cdot \frac{y}{2} = 0$,which simplifies to $-\frac{x}{2} - \frac{y}{4} = 0$,or $2x + y = 0$.The perpendicular distance from $A(1, 0)$ to the line $2x + y = 0$ is $d_1 = \frac{|2(1) + 0|}{\sqrt{2^2 + 1^2}} = \frac{2}{\sqrt{5}}$.The perpendicular distance from $B(0, 1/2)$ to the line $2x + y = 0$ is $d_2 = \frac{|2(0) + 1/2|}{\sqrt{2^2 + 1^2}} = \frac{1/2}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$.The sum of the distances is $d_1 + d_2 = \frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{4+1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$.

The straight line $x + 2y = 1$ meets the coordinate axes at $A$ and $B$. $A$ circle is drawn through $A, B$ and the origin. Then the sum of perpendicular distances from $A$ and $B$ on the tangent to the circle at the origin is

Similar Questions

The straight line $x - y - 3 = 0$ touches the circle $x^2 + y^2 - 4x + 6y + 11 = 0$ at the point whose coordinates are

The slope of the normal to the circle $x^2+y^2+2gx+2fy+c=0$ at $(x_1, y_1)$ is

$A$ is the centre of the circle $x^2+y^2-2x-4y-20=0$. If the tangents drawn at the points $B(1,7)$ and $D(4,-2)$ on the circle meet at the point $C$,then the area of the quadrilateral $ABCD$ (in square units) is

The equation of the line passing through the point $(x', y')$ and perpendicular to the line $yy' = 2a(x + x')$ is

If the line $4x - 3y + p = 0$ $(p + 3 > 0)$ touches the circle $x^2 + y^2 - 4x + 6y + 4 = 0$ at the point $(h, k)$,then $h - 2k = . . . . . .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module