Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
414 views
in Circles by (82.5k points)
closed by
The locus of the point of intersection of perpendicular tangents to the circles `x^(2)+y^(2)=a^(2)` and `x^(2)+y^(2)=b^(2)` , is
A. `x^(2)+y^(2)=a^(2)-b^(2)`
B. `x^(2)+y^(2)=a^(2)+b^(2)`
C. `x^(2)+y^(2)=(a+b)^(2)`
D. none of these

1 Answer

0 votes
by (88.1k points)
selected by
 
Best answer
Correct Answer - B
The equation of any tangent to `x^(2)+y^(2)=a^(2)` is
`x cos alpha + y sin alpha =a " " ...(i)`
The equation of the tangent to `x^(2)+y^(2)=b^(2)`, perpendicular to (i) is
`x sin alpha - y sin alpha = b " " ...(ii)`
Let (h, k) be the point of intersection of (i) and (ii). Then
`h cos alpha + k sin alpha = a " " ...(iii)`
and , ` h sin alpha- k cos alpha = b " " ...(iv)`
Squaring and adding (iii) and (iv), we get
`h^(2)+k^(2)=a^(2)+b^(2)`
Hence, the locus of (h, k) is `x^(2)+y^(2)=a^(2)+b^(2)`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...