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
247 views
in Mathematics by (91.8k points)
closed by
Find the equation of pair of tangenst drawn to circle `x^(2)+y^(2)-2x+4y-4=0` from point P(-2,3). Also find the angle between tangest.

1 Answer

0 votes
by (93.2k points)
selected by
 
Best answer
We have circle `x^(2)+y^(2)-2x+4y-4=0` or `S=0`.
Tangents are drawn from point P(-2,3) to the circle.
Equation of pair of tangents is
`T^(2)=S S_(1)`
or `(-2x+3y-(x-2)+2(y+3)-4)^(2)`
`=(x^(2)+y^(2)-2x+4y-4)((-2)^(2)+3^(2)-2(-2)+4(3)-4`
`implies (-3x+5y+4)^(2)=25(x^(2)+y^(2)-2x+4y-4)`
`implies 16x^(2)+30xy-26x+60y-116=0`
`implies 8x^(2)+15xy-13x+30y-58=0`
Here, a=8, b=0 and `h=15//2`
If angle between tangents is `theta`, then
`tan theta=(2 sqrt(h^(2)-ab))/(|a+b|)=(2sqrt((225)/(4)-0))/(|8+0|)=(15)/(8)`
`:. theta =tan ^(-1).(15)/(8)`

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

...