
A circle is drawn to cut a chord of length 2a units X-axis and to touch the Y-axis. Find the locus of the center of the circle.
A.${x^2} + {y^2} = {a^2}$
B. ${x^2} - {y^2} = {a^2}$
C.$x + y = {a^2}$
D.${x^2} - {y^2} = 4{a^2}$
Answer
160.8k+ views
Hint: First draw a rough diagram to understand the given facts. Then suppose the coordinate of the center and apply Pythagoras theorem in the right-angle triangle of the diagram to obtain the required locus.
Formula Used:
Pythagoras theorem of right-angle triangle is,
${h^2} = {a^2} + {b^2}$ , where h is the hypotenuse, a is the height and b is the base.
Complete step by step solution:
The diagram of the given problem is,

Image: Circle
Suppose that the coordinate of the circle is $C(h,k)$ .
We know that $AB = \dfrac{1}{2}AD$
Given that, $AD = 2a$ , therefore, $AB = a$ .
And, CE and CA are the radius of the circle, so the lengths are equal and equal to h.
Now, apply Pythagoras theorem in the right-angle triangle ABC,
${h^2} = {k^2} + {a^2}$
${h^2} - {k^2} = {a^2}$
Hence, the locus is,
${x^2} - {y^2} = {a^2}$
Option ‘B’ is correct
Note: The circle touches the y-axis, it means the abscissa of the center is the radius of the circle. The perpendicular from the center on a chord of a circle bisects the chord. Now apply Pythagorean theorem to find the locus of the center.
Formula Used:
Pythagoras theorem of right-angle triangle is,
${h^2} = {a^2} + {b^2}$ , where h is the hypotenuse, a is the height and b is the base.
Complete step by step solution:
The diagram of the given problem is,

Image: Circle
Suppose that the coordinate of the circle is $C(h,k)$ .
We know that $AB = \dfrac{1}{2}AD$
Given that, $AD = 2a$ , therefore, $AB = a$ .
And, CE and CA are the radius of the circle, so the lengths are equal and equal to h.
Now, apply Pythagoras theorem in the right-angle triangle ABC,
${h^2} = {k^2} + {a^2}$
${h^2} - {k^2} = {a^2}$
Hence, the locus is,
${x^2} - {y^2} = {a^2}$
Option ‘B’ is correct
Note: The circle touches the y-axis, it means the abscissa of the center is the radius of the circle. The perpendicular from the center on a chord of a circle bisects the chord. Now apply Pythagorean theorem to find the locus of the center.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
