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** **__Solve this :__ 27. __ Here another point means respect to the the given point such that the diameter comes out to be a straight line or we can directly apply section formula? __

26. In what ratio does the point (-4, 6) divide the line segment joining the point A (-6, 10) and B (3, -8) ?

27. The centre C, of a circle has the coordinates (4, 5) and one point on the circumference is (8, 10). Find the coordinates of the other end of the diameter of the circle through this point.

__Solve this :__

26. In what ratio does the point (-4, 6) divide the line segment joining the point A (-6, 10) and B (3, -8) ?

27. The centre C, of a circle has the coordinates (4, 5) and one point on the circumference is (8, 10). Find the coordinates of the other end of the diameter of the circle through this point.

Please find below the solution to the asked query:

26 )

27 ) To get the coordinates of the other end of the diameter of the circle through this point , We use mid point formula as we know center is mid point of diameter of circle .

We know formula : (

*x*,

*y*) = $\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{\displaystyle {y}_{1}+{y}_{2}}}{{\displaystyle 2}}\right)$ , Here from given information we have :

*x*= 4 ,

*y*= 5 and

*x*

_{1}= 8 and

*y*

_{1}= 10 and we assume coordinates of the other end of the diameter of the circle through this point (

*x*

_{2},

*y*

_{2}) , So

$\left(4,5\right)=\left(\frac{8+{x}_{2}}{2},\frac{{\displaystyle 10+{y}_{2}}}{{\displaystyle 2}}\right),\mathrm{So}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{8+{x}_{2}}{2}=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 8+{x}_{2}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{{\displaystyle 10+{y}_{2}}}{{\displaystyle 2}}=5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 10+{y}_{2}=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left({\mathbf{x}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{}\mathbf{)}$

Therefore,

**The coordinates of the other end of the diameter of the circle through this point = ( 0 , 0 ) ( Ans )**

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