Question

Equation of the line passing through the mid-point of intercepts made by the circle $x^2+y^2-17x-16y+60=0$ is

• A. $17x-16y=136$
• B. $16x+17y=136$
• C. $17x+16y=136$
• D. $16x-17y=136$

Answer 1

The correct option is B

$16x+17y=136$

Method 1:

we will first find the points where the given circle intersect the coordinate axes .To find those points ,we

will put y=0 and x=0 seperately.

To find the x-coordinate

$x^2+y^2-17x-16y+60=0$ is the equation of given circle.

when the circle intersects the x-axis ,y-coordinate will be zero.

$\Rightarrow x^2-17x+60=0$

$\Rightarrow (x-12)(x-5)=0$

$\Rightarrow x=5orx=12$

so the points where it intersects x-axis are (5,0) and (12,0)

$\Rightarrow$ The mid-point $\equiv (\frac{12+5}{2},0)$

=$(\frac{17}{2},0)$

To find the y-coordinate

when the circle intersects the y -axis ,the x-coordinate will be zero.

$\Rightarrow y^2-16y+60=0$

$\Rightarrow (y-10)(y-6)=0$

$\Rightarrow y=10ory=6$

so the points are (0,10) and (0,6) and the mid-point is $(0,\frac{10+6}{2})\equiv (0,8)$ we found that the mid-points

are $(\frac{17}{2},0)$ and (0,8).we will now find the equation of the straight line passing these points

$\Rightarrow \frac{x}{\frac{17}{2}}+\frac{y}{8}=1or\frac{2x}{17}+\frac{y}{8}=1$

$\Rightarrow 16x+17y=136$