the equation of the line passing through point (-2,11) and touching the circle x2+y2=25

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Please find below the solution to the asked query:

Let line be mx-y+c=0As it passes through -2,11, hence -2m-11+c=0c=2m+11As line also touches the circle, hence perpendicular distance of line from the centre of the circle be equal to radius of the circle.x2+y2=25x-02+y-02=52Centre=0,0Radius=50+0+c1+m2=5c2=251+m2As c=2m+112m+112=25+25m24m2+121+44m=25+25m225+25m2-4m2-121-44m=021m2-44m-96=021m2+28m-72m-96=07m3m+4-243m+4=07m-243m+4=0Either m=247 or m=-43When m=247c=2.247+11=48+777c=1257Hence equation of line will be247x-y+1257=024x-7y+125=0When  m=-43c=2.-43+11=-4+333c=293Hence equation of line will be-43x-y+293=04x+3y-29=0

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