Question
Download Solution PDFबिंदू X वर दोन वर्तुळे एकमेकांना बाहेरून स्पर्श करतात. बिंदू P आणि बिंदू Q वरील वर्तुळांना स्पर्श करणार्या दोन्ही वर्तुळांसाठी PQ ही एक साधी सामाईक स्पर्शिका आहे. जर वर्तुळांची त्रिज्या R आणि r असेल, तर PQ2 शोधा.
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आपल्याला माहीत आहे,
थेट सामान्य स्पर्शिकेची लांबी = √[d2 - (R - r)2]
जेथे d हे केंद्रांमधील अंतर आहे आणि R आणि r या वर्तुळांच्या त्रिज्या आहेत.
PQ = √[(R + r)2 - (R - r)2]
⇒ PQ = √[R2 + r2 + 2Rr - (R2 + r2 - 2Rr)]
⇒ PQ = √4Rr
⇒ PQ2 = 4Rr
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