Consider the circle x^2+y^2-4x-2y+c=0 whose c | vedclass

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

Question

(D) The given equation of the circle is $x^2+y^2-4x-2y+c=0$ with centre $A(2,1)$.First,we calculate the distance $AP$:$AP = \sqrt{(10-2)^2 + (7-1)^2}

Vedclass · Accepted Answer

$-20$

Vedclass · Answer

(D) The given equation of the circle is $x^2+y^2-4x-2y+c=0$ with centre $A(2,1)$.First,we calculate the distance $AP$:$AP = \sqrt{(10-2)^2 + (7-1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10$.Since $Q$ lies on the line segment $PA$ and $PQ=5$,the distance $AQ$ is:$AQ = AP - PQ = 10 - 5 = 5$.Since $AQ$ is the radius of the circle,the radius $r = 5$.The equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$,where $(h,k)$ is the centre.Substituting the values,we get $(x-2)^2 + (y-1)^2 = 5^2$.Expanding this,we get $x^2 - 4x + 4 + y^2 - 2y + 1 = 25$,which simplifies to $x^2 + y^2 - 4x - 2y - 20 = 0$.Comparing this with the given equation $x^2 + y^2 - 4x - 2y + c = 0$,we find $c = -20$.

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle at $Q$ with $PQ=5$,then $c$ is equal to

Similar Questions

$A$ black body has a maximum wavelength ${\lambda _m}$ at a temperature of $2000 \; K$. Its corresponding wavelength at a temperature of $3000 \; K$ will be:

When a body of mass $1.0 ~kg$ is suspended from a certain light spring hanging vertically,its length increases by $5 ~cm$. By suspending a $2.0 ~kg$ block to the spring,if the block is pulled through $10 ~cm$ and released,the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 ~m/s^2$)

$\frac{\tan 80^{\circ}-\tan 10^{\circ}}{\tan 70^{\circ}}$ is equal to

The chemical entities present in the thermosphere of the atmosphere are:

Jumping genes are found in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module