1. Find the equation of the cirlce whose centre is (3,-1) and which cuts off an intercept of length 6 from the line 2x - 5y + 18 = 0
first of all what is meant by "which cuts off an intercept of length 6 from the the line 2x - 5y + 18 = 0
Please explain...!!
1the circle will intersect the line at 2 points.. the distance between those two points is called the intercept
1i am getting
Circle equation as :
(x-3)2+(y+1)2 =449
please verify
1Attempt
Circle equation: (x-3)2 + (y+1)2 = k2
k has to be found out
now from line equation :
x=5y-182
put the value of x as 5y-182 in first equation
\left( \frac{5y-18}{2}-3 \right)^2+\left( y+1\right)^2=k^2 \\ \\ 29y^2-232y+\left(580-k^2 \right)=0 \\ \texttt{Hence we can infer}\\ y_1+y_2=8\\ y_1.y_2=\frac{580-k^2}{29} \\ \\\texttt{Since} \ x= \frac{5y-18}{2} \\ \\ x_1+x_2=2 \\ \\ x_1.x_2=\left(\frac{25}{4.29}\left(580-k^2 \right)-99 \right) \\ \texttt{Intercept length is 6 ,so} \\ (x_1-x_2)^2+(y_1-y_2)^2=36 \\ \texttt{Using .... and putting in the above equation}\\ (x_1-x_2)^2=(x_1+x_2)^2-4x_1.x_2 \\ (y_1-y_2)^2=(y_1+y_2)^2-4y_1.y_2 \\ \texttt{We finally get }k^2=449
36got the solution from 1 of my friends...
on drawing the perpndicular from center to chord helps us calculate the radius nd
with the known centre nd radius the eqn can be formed...lution from 1 of my friends...
on drawing the perpndicular from center to chord helps us calculate the radius nd
with the known centre nd radius the eqn can be formed...
