The lines joining the points of intersection | vedclass

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

Question

(B) The equation of the circle is $x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - c^2 = 0$. The equation of the line is $\frac{kx + hy}{2hk} = 1$,which is $\frac

Vedclass · Accepted Answer

$c^2 = h^2 + k^2$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - c^2 = 0$. The equation of the line is $\frac{kx + hy}{2hk} = 1$,which is $\frac{x}{2h} + \frac{y}{2k} = 1$. Making the circle equation homogeneous using the line equation: $(x^2 + y^2) - 2(hx + ky)\left(\frac{x}{2h} + \frac{y}{2k}ight) + (h^2 + k^2 - c^2)\left(\frac{x}{2h} + \frac{y}{2k}ight)^2 = 0$. Expanding this,the coefficient of $x^2$ is $A = 1 - 1 + \frac{h^2 + k^2 - c^2}{4h^2} = \frac{h^2 + k^2 - c^2}{4h^2}$. The coefficient of $y^2$ is $B = 1 - 1 + \frac{h^2 + k^2 - c^2}{4k^2} = \frac{h^2 + k^2 - c^2}{4k^2}$. For the lines to be perpendicular,$A + B = 0$. $(h^2 + k^2 - c^2) \left(\frac{1}{4h^2} + \frac{1}{4k^2}ight) = 0$. Since $\frac{1}{4h^2} + \frac{1}{4k^2} 
eq 0$,we must have $h^2 + k^2 - c^2 = 0$,or $c^2 = h^2 + k^2$.

The lines joining the points of intersection of the curve $(x - h)^2 + (y - k)^2 - c^2 = 0$ and the line $kx + hy = 2hk$ to the origin are perpendicular,then

Similar Questions

The angle between the lines joining the origin to the points of intersection of the curves $2x^2 + 3y^2 + 10x = 0$ and $3x^2 + 5y^2 + 16x = 0$ is

The equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the straight line $x+y+2=0$ is

The line $x+y=k$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k (>1)$ is

The lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 3$ and the line $x + y = 2$ are:

The distance between the parallel lines given by $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module