Find the equation of the normals to the circle `x^2+y^2-8x-2y+12=0`
at the point whose ordinate is `-1`
Text Solution
Verified by Experts
We have circle `x^(2)+y^(2)-8x-2y+12=0` Centre of the circle is C(4,1). Putting `y= -1` in the equation of the circle, we get `x^(2)-8x+15=0` `implies(x-3)(x-5)=0` `implies x=5 ` or 3 Thus, the points on the circle are P(5,-1) and Q(3,-1). Equation of normal at P is `y+1=(-1-1)/(5-4)(x-5)` or `2x+y-9=0` Equation of normal at Q is `y+1=(-1-1)/(3-4)(x-5)` or `2x-y-7=0`
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