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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. |
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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)` |
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