The equation for QR is 5y = – 4x +41. Is QR tangent to Circle O at R?<\/p>\n\n\n\n
A. No, because the slope of OR times the slope of QR does not equal -1. YA Q R(4.5) Yes, because the slope of OR times the slope of QR equals<\/p>\n\n\n\n
B. No, because the slope of OR times the slope of QR does not equal 1. Yes, because the slope of OR times the slope of QR equals 1<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n A. No, because the slope of OR times the slope of QR does not equal -1.<\/strong><\/p>\n\n\n\n To determine whether the line QRQR is tangent to the circle OO at point RR, we must verify whether QRQR is perpendicular to the radius OROR at RR. For perpendicularity, the product of the slopes of OROR and QRQR must equal \u22121-1. Here\u2019s the step-by-step reasoning:<\/p>\n\n\n\n The equation for QR is 5y = – 4x +41. Is QR tangent to Circle O at R? A. No, because the slope of OR times the slope of QR does not equal -1. YA Q R(4.5) Yes, because the slope of OR times the slope of QR equals B. No, because the slope of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-180295","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180295","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=180295"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180295\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180295"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180295"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180295"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Correct Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
\n\n\n\n\n
The equation of QRQR is given as 5y=\u22124x+415y = -4x + 41.
Rewrite it in slope-intercept form y=mx+by = mx + b: y=\u221245x+415y = -\\frac{4}{5}x + \\frac{41}{5} The slope of QRQR is mQR=\u221245m_{QR} = -\\frac{4}{5}.<\/li>\n<\/ol>\n\n\n\n
\n\n\n\n\n
Let RR be a point on the circle where QRQR intersects it. Assume the circle OO has its center at (h,k)(h, k). The slope of OROR, denoted as mORm_{OR}, is calculated using the coordinates of O(h,k)O(h, k) and R(x1,y1)R(x_1, y_1). mOR=y1\u2212kx1\u2212hm_{OR} = \\frac{y_1 – k}{x_1 – h} Without specific coordinates for OO, h,k,x1,h, k, x_1, or y1y_1, the product of slopes mOR\u22c5mQRm_{OR} \\cdot m_{QR} cannot be definitively calculated at this step.<\/li>\n<\/ol>\n\n\n\n
\n\n\n\n\n
A line is tangent to a circle if and only if it is perpendicular to the radius at the point of tangency. This condition requires: mOR\u22c5mQR=\u22121m_{OR} \\cdot m_{QR} = -1 If this condition is not satisfied, QRQR is not<\/strong> tangent to the circle.<\/li>\n<\/ol>\n\n\n\n
\n\n\n\n\n
If mOR\u22c5mQR\u2260\u22121m_{OR} \\cdot m_{QR} \\neq -1, the line QRQR is not tangent to the circle OO at RR. Thus, the correct answer is A<\/strong>.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"