The circle {x^2} + {y^2} = 4 cuts the line jo | vedclass

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

Question

(A) The equation of the line passing through $A(1, 0)$ and $B(3, 4)$ is given by $y - 0 = \frac{4 - 0}{3 - 1}(x - 1)$,which simplifies to $y = 2x - 2$

Vedclass · Accepted Answer

$3{x^2} + 2x - 21 = 0$

Vedclass · Answer

(A) The equation of the line passing through $A(1, 0)$ and $B(3, 4)$ is given by $y - 0 = \frac{4 - 0}{3 - 1}(x - 1)$,which simplifies to $y = 2x - 2$.Substituting $y = 2x - 2$ into the circle equation ${x^2} + {y^2} = 4$:${x^2} + {(2x - 2)^2} = 4$${x^2} + 4{x^2} - 8x + 4 = 4$$5{x^2} - 8x = 0$$x(5x - 8) = 0$,so $x = 0$ or $x = \frac{8}{5}$.For $x = 0$,$y = 2(0) - 2 = -2$. Thus,$Q = (0, -2)$.For $x = \frac{8}{5}$,$y = 2(\frac{8}{5}) - 2 = \frac{16}{5} - \frac{10}{5} = \frac{6}{5}$. Thus,$P = (\frac{8}{5}, \frac{6}{5})$.Now,calculate the ratios using the section formula or distance formula. Since $P$ and $Q$ lie on the segment $AB$ or its extension,we use the ratio of distances.$A = (1, 0)$,$B = (3, 4)$,$P = (1.6, 1.2)$,$Q = (0, -2)$.$PA = \sqrt{(1.6 - 1)^2 + (1.2 - 0)^2} = \sqrt{0.6^2 + 1.2^2} = \sqrt{0.36 + 1.44} = \sqrt{1.8} = \frac{3}{\sqrt{5}}$.$BP = \sqrt{(3 - 1.6)^2 + (4 - 1.2)^2} = \sqrt{1.4^2 + 2.8^2} = \sqrt{1.96 + 7.84} = \sqrt{9.8} = \frac{7}{\sqrt{5}}$.$\alpha = \frac{BP}{PA} = \frac{7/\sqrt{5}}{3/\sqrt{5}} = \frac{7}{3}$.$QA = \sqrt{(1 - 0)^2 + (0 - (-2))^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$.$BQ = \sqrt{(3 - 0)^2 + (4 - (-2))^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.Since $Q$ divides $AB$ externally,$\beta = \frac{BQ}{QA} = \frac{3\sqrt{5}}{-\sqrt{5}} = -3$.The quadratic equation with roots $\alpha = \frac{7}{3}$ and $\beta = -3$ is:${x^2} - (\alpha + \beta)x + \alpha\beta = 0$${x^2} - (\frac{7}{3} - 3)x + (\frac{7}{3})(-3) = 0$${x^2} - (-\frac{2}{3})x - 7 = 0$${x^2} + \frac{2}{3}x - 7 = 0$$3{x^2} + 2x - 21 = 0$.

The circle ${x^2} + {y^2} = 4$ cuts the line joining the points $A(1, 0)$ and $B(3, 4)$ in two points $P$ and $Q$. Let $\frac{BP}{PA} = \alpha$ and $\frac{BQ}{QA} = \beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation

Similar Questions

Let $y=mx+c, m>0$ be the focal chord of $y^{2}=-64x$,which is tangent to $(x+10)^{2}+y^{2}=4$. Then,the value of $4\sqrt{2}(m+c)$ is equal to $.....$

If the length of the tangent from any point on the circle $(x-3)^2+(y+2)^2=5r^2$ to the circle $(x-3)^2+(y+2)^2=r^2$ is $16$ units, then the area between the two circles in sq. units is (in $\pi$)

Consider the two curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Then,

If the portion of the line $lx + my = 1$ falling inside the circle ${x^2} + {y^2} = {a^2}$ subtends an angle of $45^\circ$ at the origin,then

Let the point $P$ be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points $R$ and $S$, then the maximum value of $(PR + PS)^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module