Question

find the mid point of the chord intercept by x^2+y^2-2x-10y+1=0 on the line x-2y+7=0​

Answer 1

Answer:

• pls mark as brainliest friend

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

The given equation of the circle is

$x^2+y^2-2x-10y+1=0 \: \: \: ....(1)$

The given equation of the line is

$x-2y+7=0 \: \: \: .........(2)$

For points of intersection of (1) & (2) we put x = 2y - 7 in Equation (1)

$(2y - 7)^2+y^2-2 \times (2y - 7)-10y+1=0$

$\implies \: 4 {y}^{2} - 28y + 49 + {y}^{2} - 4y + 14 - 10y + 1 = 0$

$\implies \: 5{y}^{2} - 42y + 64= 0$

$\implies \: 5 {y}^{2} - 32y - 10y + 64 = 0$

$\implies \: y(5 {y} - 32) - 2(5y - 32) = 0$

$\implies \: (y - 2)(5 {y} - 32) = 0$

Which gives

$y = 2 \: \: and \: \: y = \frac{32}{5}$

Now

$y = 2 \: \: implies \: \: x = 4 - 7 = - 3$

$y = \frac{32}{5} \: \: implies \: \: x = \frac{29}{5}$

So the two extremities of the chord are

$( - 3, 2) \: \: and \: \: \: ( \frac{29}{5}, \frac{32}{5} )$

So the required midpoint is

$\displaystyle \: (\frac{ - 3 + \frac{29}{5} }{2} , \frac{2 + \frac{32}{5} }{2} ) = ( \frac{7}{5} , \frac{21}{5} )$