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Question: Prove that \( (20,3),(19,8),(2, - 9)\& (2,15) \) are concyclic. Also find the equation of the circle...

Prove that (20,3),(19,8),(2,−9)&(2,15)(20,3),(19,8),(2, - 9)\& (2,15) are concyclic. Also find the equation of the circle passing through them.

Explanation

Solution

Hint : We should first consider that three points are collinear. Using this assumption we will find an equation of the circle. Then we substitute the value of the last point into the equation of the triangle and check if that satisfies the equation of the circle.

Complete step by step solution:
Let the points be A(20,3),B(19,8),C(2,−9),D(2,15)A(20,3),B(19,8),C(2, - 9),D(2,15) .
We know that the standard equation of circle is :
x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0
Now, the equation of circle passing through B(19,8)B(19,8) is:
(19)2+(8)2+2g(19)+2f(8)+c=0 361+64+38g+16f+c=0 425+38g+16f+c=0−(i)   {(19)^2} + {(8)^2} + 2g(19) + 2f(8) + c = 0 \\\ 361 + 64 + 38g + 16f + c = 0 \\\ 425 + 38g + 16f + c = 0 - (i) \;
The equation of circle passing through C(2,−9)C(2, - 9) is:
(2)2+(−9)2+2g(2)+2f(−9)+c=0 4+81+4g−18f+c=0 85+4g−18f+c=0−(ii)   {(2)^2} + {( - 9)^2} + 2g(2) + 2f( - 9) + c = 0 \\\ 4 + 81 + 4g - 18f + c = 0 \\\ 85 + 4g - 18f + c = 0 - (ii) \;
The equation of circle passing through D(2,15)D(2,15) is:
(2)2+(15)2+2g(2)+2f(15)+c=0 4+225+4g+30f+c=0 229+4g+30f+c=0−(iii)   {(2)^2} + {(15)^2} + 2g(2) + 2f(15) + c = 0 \\\ 4 + 225 + 4g + 30f + c = 0 \\\ 229 + 4g + 30f + c = 0 - (iii) \;
Now, when we subtract (ii)(ii) from (iii)(iii) , we get,
229+4g+30f+c−85−4g+18f−c=0 229+4g+30f+c−85−4g+18f−c=0 144+48f=0 144=−48f f=−14448 f=−3   229 + 4g + 30f + c - 85 - 4g + 18f - c = 0 \\\ 229 + 4{g} + 30f +{c} - 85 - 4{g} + 18f -{c} = 0 \\\ 144 + 48f = 0 \\\ 144 = - 48f \\\ f = \dfrac{{ - 144}}{{48}} \\\ f = - 3 \;
Now subtracting (i)(i) from (ii)(ii) , we get,
425+38g+16f+c−229−4g−30f−c=0 425+38g+16f+c−229−4g−30f−c=0 196+34g−14f=0   425 + 38g + 16f + c - 229 - 4g - 30f - c = 0 \\\ 425 + 38g + 16f +{c} - 229 - 4g - 30f -{c} = 0 \\\ 196 + 34g - 14f = 0 \;
Now, we know that f=−3f = - 3
196+34g−14(−3)=0 196+42+34g=0 238+34g=0 g=−23834=−7   196 + 34g - 14( - 3) = 0 \\\ 196 + 42 + 34g = 0 \\\ 238 + 34g = 0 \\\ g = \dfrac{{ - 238}}{{34}} = - 7 \;
Substituting the values to (i)(i) we get,
425+38(−7)+16(−3)+c=0 425−266−48+c=0 c=−111   425 + 38( - 7) + 16( - 3) + c = 0 \\\ 425 - 266 - 48 + c = 0 \\\ c = - 111 \;
Hence, the equation of the circle is x2+y2−14x−6y−111=0{x^2} + {y^2} - 14x - 6y - 111 = 0 .
Now, when we substitute the values of point A(20,3)A(20,3) into the equation of circle, we get,
x2+y2−14x−6y−111=0{x^2} + {y^2} - 14x - 6y - 111 = 0
(20)2+(3)2−14(20)−6(3)−111=0 400+9−280−18−111=0 409−409=0 0=0   {(20)^2} + {(3)^2} - 14(20) - 6(3) - 111 = 0 \\\ 400 + 9 - 280 - 18 - 111 = 0 \\\ 409 - 409 = 0 \\\ 0 = 0 \;
Now, since the left hand side is equal to the right hand side .
Hence, it satisfies the equation of circle. Therefore, It is also concyclic.

Note : A set of points is said to be concyclic only when all the points lie on a common circle. The distance between all the points and the centre of the circle will remain the same.