Question

The midpoint of the chord  2x+y-5=0 with respect to the circle x²+y²=9 is​

Answer 1

Answer:

38

Step-by-step explanation:

Answer 2

Answer:

Midpoint of the chord with respect to the circle is (2, 1)

Step-by-step explanation:

Given, a circle $x^{2} +y^{2} =9$

The standard equation of a circle is $x^{2} +y^{2} =r^{2}$

Comparing both the equations, $r^{2}=9\implies r=3$

so, radius of the circle is 3units.

Given, the equation for chord is $2x+y-5=0$

Let (h, k) be the midpoint of the chord.

Since the point is on the chord, substituting h and k in place of x and y in chord equation, we get $2h+k-5=0$                       ....(1)

From the figure, it can be seen that OP is perpendicular to the chord

Therefore the slope equation is $m_{1} m_{2} =-1$             ....(2)

where $m_{1}$ and $m_{2}$ are the slopes of chord and OP.  

Slope of the  chord is  $m_{1} =-\frac{2}{1}=-2$

Slope of OP is $m_{2} =\frac{y_{2}- y_{1}}{{x_{2}- x_{1}}}$

                              $=\frac{k- 0}{h- 0}=\frac{k}{h}$

Substituting the slope in equation (2), we get

                                     $-2. \frac{k}{h} =-1\implies h=2k$

Substituting h inequation (1), $2(2k)+k-5=0\implies 5k=5\implies k=1$

and hence $h=2k=2\times 1=2$

Hence, the midpoint of the chord with respect to the circle is (2, 1)