Question 45
NECO
•
2021
•
FURTHER_MATHEMATICS
•
objective
Find the equation of the circle that passes through the point \(A(2,3)\), is tangent to the line \(y = x + 1\), and whose centre lies on the line \(2x - y = 0\). Determine the radius of the circle.
Answer Options
A.
Centre \\left(\\frac{4}{3},\\frac{8}{3}\\right); radius \\frac{\\sqrt{2}}{3}; equation \\left(x-\\frac{4}{3}\\right)^{2}+\\left(y-\\frac{8}{3}\\right)^{2}=\\frac{2}{9}
B.
Centre \\left(\\frac{5}{3},\\frac{10}{3}\\right); radius \\frac{\\sqrt{5}}{3}; equation \\left(x-\\frac{5}{3}\\right)^{2}+\\left(y-\\frac{10}{3}\\right)^{2}=\\frac{5}{9}
C.
Centre \\left(\\frac{5}{3},\\frac{10}{3}\\right); radius \\frac{\\sqrt{2}}{3}; equation \\left(x-\\frac{5}{3}\\right)^{2}+\\left(y-\\frac{10}{3}\\right)^{2}=\\frac{4}{9}
Correct Answer
D.
Centre \left(\frac{5}{3},\frac{10}{3}\right); radius \frac{\sqrt{2}}{3}; equation \left(x-\frac{5}{3}\right)^{2}+\left(y-\frac{10}{3}\right)^{2}=\frac{2}{9}
Correct Answer
D
Correct Option:
Centre \left(\frac{5}{3},\frac{10}{3}\right); radius \frac{\sqrt{2}}{3}; equation \left(x-\frac{5}{3}\right)^{2}+\left(y-\frac{10}{3}\right)^{2}=\frac{2}{9}
About This Question
This is
Neco
2021
Further Mathematics
Question 45.
It is one of the
objective
questions from the
2021
Neco
Further Mathematics
examination.
Difficulty level:
Medium
.