MHT CET · Maths · Circle
If a circle passes through points \((4,0)\) and \((0,2)\) and its centre lies on \(\mathrm{Y}\)-axis. If the radius of the circle is \(r\), then the value of \(r^2-r+1\) is
- A \(25\)
- B \(21\)
- C \(20\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(21\)
Step-by-step Solution
Detailed explanation
Let \((0, y)\) be the centre of the circle.
\(\therefore \quad \sqrt{(0-4)^2+(y-0)^2}=\sqrt{(0-0)^2+(y-2)^2} \)
\( \therefore \quad 16+y^2=(y-2)^2 \)
\( \therefore \quad 16+y^2=y^2-4 y+4 \)
\( \therefore y=-3 \)
\( \therefore \text { centre of the circle is }(0,-3) . \)
\( \therefore \text { Radius of the circle }=\mathrm{r}=\) \(\sqrt{(0-0)^2+(-3-2)^2} \)
\( =5 \text { units } \)
\( \therefore \mathrm{r}^2-\mathrm{r}+1=25-5+1=21\)
\(\therefore \quad \sqrt{(0-4)^2+(y-0)^2}=\sqrt{(0-0)^2+(y-2)^2} \)
\( \therefore \quad 16+y^2=(y-2)^2 \)
\( \therefore \quad 16+y^2=y^2-4 y+4 \)
\( \therefore y=-3 \)
\( \therefore \text { centre of the circle is }(0,-3) . \)
\( \therefore \text { Radius of the circle }=\mathrm{r}=\) \(\sqrt{(0-0)^2+(-3-2)^2} \)
\( =5 \text { units } \)
\( \therefore \mathrm{r}^2-\mathrm{r}+1=25-5+1=21\)
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