Let A(2, 3), B(4, 5) be two points and let C | vedclass

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

Question

(D) The given equation $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$ represents a circle with diameter $AB$,where $A = (2, 3)$ and $B = (4, 5)$.The length of

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The given equation $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$ represents a circle with diameter $AB$,where $A = (2, 3)$ and $B = (4, 5)$.The length of the diameter $AB = \sqrt{(4 - 2)^2 + (5 - 3)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes h = \sqrt{2}$.Substituting $AB = 2\sqrt{2}$,we get $\frac{1}{2} 	imes 2\sqrt{2} 	imes h = \sqrt{2}$,which simplifies to $\sqrt{2} 	imes h = \sqrt{2}$,so $h = 1$.The height $h$ is the perpendicular distance from point $C$ to the line segment $AB$. Since the radius of the circle is $r = \frac{AB}{2} = \sqrt{2} \approx 1.414$,and $h = 1 Each of these two lines intersects the circle at two distinct points. Therefore,there are $4$ possible positions for point $C$.

Let $A(2, 3), B(4, 5)$ be two points and let $C = (x, y)$ be a point such that $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$. If the area of $\Delta ABC = \sqrt{2}$ sq. unit,then the maximum number of positions of $C$ in the $xy$ plane is

Similar Questions

Given $r_1, r_2 > 0$ and $C_1, C_2$ are the centres of two circles having only two common tangents. If $C_1 C_2 = r_1 + r_2$,which of the following is correct?

The length of the chord joining points $(4 \cos \theta, 4 \sin \theta)$ and $(4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$ on the circle $x^2+y^2=16$ is

If the straight line $y = mx$ is outside the circle $x^2 + y^2 - 20y + 90 = 0$,then

The two circles $x^2 + y^2 - 4y = 0$ and $x^2 + y^2 - 8y = 0$:

The length of the transverse common tangent of the circles $x^2+y^2-2x+4y+4=0$ and $x^2+y^2+4x-2y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module