Solveeit Logo
Login

Question

Question: Show that the points \[\left( {5,5} \right),\left( {6,4} \right),\left( { - 2,4} \right)\& \left( {7...

Show that the points (5,5),(6,4),(−2,4)&(7,1)\left( {5,5} \right),\left( {6,4} \right),\left( { - 2,4} \right)\& \left( {7,1} \right) are concyclic. Find its equation, center and radius.

Explanation

Solution

Hint : Given are four points. That needs to be shown concyclic that is, they lie in the same circle. For that purpose we will consider three points that satisfy the standard equation of a circle. From that we will get three equations. From those equations we will find the value of g and f points of the circle and will find the actual equation of the circle.
Then we will check whether the last point satisfies the equation or not. If it satisfies then all the points are concyclic. Then getting the center and radius is very easy.

Complete step-by-step answer :
The given points are (5,5),(6,4),(−2,4)&(7,1)\left( {5,5} \right),\left( {6,4} \right),\left( { - 2,4} \right)\& \left( {7,1} \right)
Let the general equation of a circle is x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0→equation1
Now we will consider the three points as (5,5),(6,4)&(7,1)\left( {5,5} \right),\left( {6,4} \right)\& \left( {7,1} \right)
Suppose they satisfy the equation of the circle. So putting the values of the points one by one in the equation1 we get,
For the point (5,5)\left( {5,5} \right); 52+52+2×5g+2×5f+c=0{5^2} + {5^2} + 2 \times 5g + 2 \times 5f + c = 0
Taking the square and multiplying the numbers,
25+25+10g+10f+c=025 + 25 + 10g + 10f + c = 0
On adding, 50+10g+10f+c=050 + 10g + 10f + c = 0 →equation2

For the point (6,4)\left( {6,4} \right); 62+42+2×6g+2×4f+c=0{6^2} + {4^2} + 2 \times 6g + 2 \times 4f + c = 0
Taking the square and multiplying the numbers,
36+16+12g+8f+c=036 + 16 + 12g + 8f + c = 0
On adding, 52+12g+8f+c=052 + 12g + 8f + c = 0 →equation3

For the point (7,1)\left( {7,1} \right); 72+12+2g×7+2f×1+c=0{7^2} + {1^2} + 2g \times 7 + 2f \times 1 + c = 0
Taking the square and multiplying the numbers,
49+1+14g+2f+c=049 + 1 + 14g + 2f + c = 0
On adding, 50+14g+2f+c=050 + 14g + 2f + c = 0 →equation4
Now in order to find the values of g and f we will use the equations above,
equation4 - equation2 will be,
50+14g+2f+c−(50+10g+10f+c)=050 + 14g + 2f + c - \left( {50 + 10g + 10f + c} \right) = 0
On multiplying the minus sign with the bracket,
50+14g+2f+c−50−10g−10f−c=050 + 14g + 2f + c - 50 - 10g - 10f - c = 0
On cancelling the opposite signed terms,
14g−10g+2f−10f=014g - 10g + 2f - 10f = 0
On solving we get,
4g−8f=04g - 8f = 0
On dividing both sides by 4 we get,
g−2f=0g - 2f = 0→equation5
equation4 – equation3 will be,
50+14g+2f+c−(52+12g+8f+c)=050 + 14g + 2f + c - \left( {52 + 12g + 8f + c} \right) = 0
On multiplying the minus sign with the bracket,
50+14g+2f+c−52−12g−8f−c=050 + 14g + 2f + c - 52 - 12g - 8f - c = 0
On cancelling the opposite signed terms,
50−52+14g−12g+2f−8f+c−c=050 - 52 + 14g - 12g + 2f - 8f + c - c = 0
On solving it we get,
−2+2g−6f=0- 2 + 2g - 6f = 0
On dividing both sides by 2 we get,
−1+g−3f=0- 1 + g - 3f = 0
Taking constants on other side we get,
g−3f=1g - 3f = 1 →equation6
Now to find the value of g and f we will,
equation5 – equation6
g−2f−(g−3f)=0−1g - 2f - \left( {g - 3f} \right) = 0 - 1
On multiplying the bracket with minus sign,
g−2f−g+3f=0−1g - 2f - g + 3f = 0 - 1
cancelling the opposite terms,
−2f+3f=−1- 2f + 3f = - 1
on adding the same terms we get,
f=−1f = - 1
now taking this value of f and putting it in equtaion6 we get,
g=1+3(−1)g = 1 + 3\left( { - 1} \right)
On multiplying with the bracket,
g=1−3g = 1 - 3
On subtracting we get,
g=−2g = - 2
Now we have the values of g and f. Now the value of c is to be found. Putting these values in equation2 we get,
50+10(−2)+10(−1)=−c50 + 10\left( { - 2} \right) + 10\left( { - 1} \right) = - c
Now on calculating we get,
50−20−10=−c50 - 20 - 10 = - c
On taking the calculations we get,
c=−20c = - 20
thus the equation of circle is,
x2+y2+2(−2)x+2(−1)y−20=0{x^2} + {y^2} + 2\left( { - 2} \right)x + 2\left( { - 1} \right)y - 20 = 0
On calculating we get,
x2+y2−4x−2y−20=0{x^2} + {y^2} - 4x - 2y - 20 = 0→equation7
now we will test this equation for the last point (−2,4)\left( { - 2,4} \right)
=(−2)2+42−4(−2)−2×4−20= {\left( { - 2} \right)^2} + {4^2} - 4\left( { - 2} \right) - 2 \times 4 - 20
On multiplying the respective brackets,
=4+16+8−8−20= 4 + 16 + 8 - 8 - 20
On calculating we get,
=20−20=0= 20 - 20 = 0
Thus it satisfies the equation. So we can say that the four points are concyclic.
The centre of the circle is, (−g,−f)=(−(−2),−(−1))=(2,1)\left( { - g, - f} \right) = \left( { - \left( { - 2} \right), - \left( { - 1} \right)} \right) = \left( {2,1} \right)
The radius is given by,
r=g2+f2−cr = \sqrt {{g^2} + {f^2} - c}
On putting the values,
r=(−2)2+(−1)2−(−20)r = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 1} \right)}^2} - \left( { - 20} \right)}
On calculating we get,
r=25r = \sqrt {25}
Taking the root we get,
r=5r = 5
Thus this completes the solution.
Equation of circle → x2+y2−4x−2y−20=0{x^2} + {y^2} - 4x - 2y - 20 = 0
Centre of the circle → (2,1)\left( {2,1} \right)
Radius of the circle → 5 units

Note : Here note that the solution is too long but simplifying the equations carefully will help to go nearer to solution. Also note that we can choose any three points. Not necessary that we should take the points we have taken in the question above. But for sure the steps should be the same