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Find the equation of the circle with radius 5 whose centre lies onxaxis and passes through the point (2, 3). |
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Answer» Let the centre of circle on x-axis be (h,0). Given that radius of circle = 5 Therefore, equation of circle is `(x-h)^(2)+(y-0)^(2)=5^(2)` `rArr""(x-h)^(2)+y^(2)=25` . . .(1) This circle passes through the point (2,3). `:." "(2-h)^(2)+3^(3)=25` `rArr" "(2-h)^(2)=16` `rArr" "2-h=pm4` `rArr" "h=6orh=-2`. `:.` From equation (1) Equation of cicle `(x-6)^(2)+y^(2)=25or(x+2)^(2)+y^(2)=25` |
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